TS EAMCET 2021 Mathematics Question Paper with Answer and Solution

(D) Let the lines be $y - mx = 0$ and $y - m_1x = 0$ for the first equation,and $y - mx = 0$ and $y - m_2x = 0$ for the second equation,where $y - mx = 0$ is the common line.
For $2x^2 + \alpha xy + 3y^2 = 0$,we have $3y^2 + \alpha xy + 2x^2 = 0$. Dividing by $x^2$,we get $3(y/x)^2 + \alpha(y/x) + 2 = 0$.
Let $m$ and $m_1$ be the roots. Then $m + m_1 = -\alpha/3$ and $m \cdot m_1 = 2/3$.
For $2x^2 + \beta xy - 3y^2 = 0$,we have $-3y^2 + \beta xy + 2x^2 = 0$,or $3y^2 - \beta xy - 2x^2 = 0$. Dividing by $x^2$,we get $3(y/x)^2 - \beta(y/x) - 2 = 0$.
Let $m$ and $m_2$ be the roots. Then $m + m_2 = \beta/3$ and $m \cdot m_2 = -2/3$.
Given that the other two lines $y - m_1x = 0$ and $y - m_2x = 0$ are perpendicular,$m_1 \cdot m_2 = -1$.
From $m \cdot m_1 = 2/3$,$m_1 = 2/(3m)$.
From $m \cdot m_2 = -2/3$,$m_2 = -2/(3m)$.
Substituting into $m_1 \cdot m_2 = -1$: $(2/(3m)) \cdot (-2/(3m)) = -1$ $\Rightarrow -4/(9m^2) = -1$ $\Rightarrow m^2 = 4/9$ $\Rightarrow m = \pm 2/3$.
If $m = 2/3$,$m_1 = 1$ and $m_2 = -1$.
Then $\alpha = -3(m + m_1) = -3(2/3 + 1) = -5$ and $\beta = 3(m + m_2) = 3(2/3 - 1) = -1$.
$|\alpha + \beta| = |-5 - 1| = 6$.
If $m = -2/3$,$m_1 = -1$ and $m_2 = 1$.
Then $\alpha = -3(-2/3 - 1) = 5$ and $\beta = 3(-2/3 + 1) = 1$.
$|\alpha + \beta| = |5 + 1| = 6$.

(B) The given line is $\frac{x}{3}+\frac{y}{2}=1$,which can be written as $2x+3y=6$,or $y=-\frac{2}{3}x+2$. The slope of this line is $m_1=-\frac{2}{3}$.
Since the pair of lines passing through the origin are perpendicular and form an isosceles triangle with the given line,the angles they make with the given line must be $45^{\circ}$.
Let the slopes of the pair of lines be $m$ and $-\frac{1}{m}$ (since they are perpendicular).
The angle between the line $y=mx$ and the line $y=-\frac{2}{3}x+2$ is $45^{\circ}$.
$\tan 45^{\circ} = \left| \frac{m - (-2/3)}{1 + m(-2/3)} \right| = 1$
$1 = \left| \frac{3m+2}{3-2m} \right|$
$3m+2 = 3-2m$ or $3m+2 = -(3-2m)$
$5m = 1 \Rightarrow m = \frac{1}{5}$ or $m+5 = 0 \Rightarrow m = -5$.
The lines are $y=\frac{1}{5}x$ and $y=-5x$,which are $x-5y=0$ and $5x+y=0$.
The joint equation is $(x-5y)(5x+y) = 0$.
$5x^2 + xy - 25xy - 5y^2 = 0$
$5x^2 - 24xy - 5y^2 = 0$
$5(x^2-y^2) - 24xy = 0$.

(B) Given the pair of lines $S \equiv 2x^2+3xy-2y^2-7x+y+3=0$.
Factorizing the homogeneous part: $2x^2+3xy-2y^2 = (x+2y)(2x-y) = 0$.
Let the lines be $(x+2y+c_1)(2x-y+c_2) = 2x^2+3xy-2y^2-7x+y+3$.
Comparing coefficients of $x$ and $y$: $2c_1+c_2 = -7$ and $2c_2-c_1 = 1$.
Solving these,we get $c_1 = -3$ and $c_2 = -1$.
So,the lines are $x+2y-3=0$ and $2x-y-1=0$.
The lines $L_1$ and $L_2$ are perpendicular to these lines and pass through $(3,-4)$.
Line perpendicular to $x+2y-3=0$ is $2x-y+k_1=0$. Passing through $(3,-4)$: $2(3)-(-4)+k_1=0 \Rightarrow k_1=-10$. So,$2x-y-10=0$.
Line perpendicular to $2x-y-1=0$ is $x+2y+k_2=0$. Passing through $(3,-4)$: $3+2(-4)+k_2=0 \Rightarrow k_2=5$. So,$x+2y+5=0$.
The area of the rectangle formed by these four lines is the product of the distances between the parallel pairs.
Distance between $x+2y-3=0$ and $x+2y+5=0$ is $d_1 = \frac{|5-(-3)|}{\sqrt{1^2+2^2}} = \frac{8}{\sqrt{5}}$.
Distance between $2x-y-1=0$ and $2x-y-10=0$ is $d_2 = \frac{|-10-(-1)|}{\sqrt{2^2+(-1)^2}} = \frac{9}{\sqrt{5}}$.
Area $= d_1 \times d_2 = \frac{8}{\sqrt{5}} \times \frac{9}{\sqrt{5}} = \frac{72}{5} \text{ sq. units}$.

(C) The bisectors of the angle between the coordinate axes are $y = x$ and $y = -x$.
Since these lines are part of the pair of lines represented by $x^2+2axy+3y^2=0$,they must satisfy the equation.
Case $1$: Substitute $y = x$ into the equation:
$x^2+2ax(x)+3x^2 = 0$
$4x^2+2ax^2 = 0$
$x^2(4+2a) = 0$
Since this holds for all $x$,$4+2a = 0$,which gives $a = -2$.
Case $2$: Substitute $y = -x$ into the equation:
$x^2+2ax(-x)+3(-x)^2 = 0$
$x^2-2ax^2+3x^2 = 0$
$4x^2-2ax^2 = 0$
$x^2(4-2a) = 0$
Since this holds for all $x$,$4-2a = 0$,which gives $a = 2$.
The possible values of $a$ are $-2$ and $2$.
Therefore,the sum of all possible values of $a$ is $(-2) + 2 = 0$.

(A) The given equation of the pair of straight lines is $3y^2 - 8xy - 3x^2 - 29x + 3y - 18 = 0$.
Rearranging the terms,we have $-3x^2 - 8xy + 3y^2 - 29x + 3y - 18 = 0$.
Comparing this with the general equation of a pair of straight lines $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a = -3$ and $b = 3$.
Since the sum of the coefficients of $x^2$ and $y^2$ is $a + b = -3 + 3 = 0$,the lines are perpendicular to each other.
Therefore,the angle between the pair of straight lines is $90^{\circ}$.

(A) The given equation of the pair of lines is $x^2 + 3y^2 + 4xy - 4x - 10y + 3 = 0$.
Comparing this with the general form $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$,we get $a = 1, b = 3, h = 2, g = -2, f = -5, c = 3$.
The product of the lengths of the perpendiculars from the origin $(0, 0)$ to the pair of lines represented by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is given by the formula $\frac{|c|}{\sqrt{(a-b)^2 + (2h)^2}}$.
Substituting the values,we get $\frac{|3|}{\sqrt{(1-3)^2 + (2 \times 2)^2}} = \frac{3}{\sqrt{(-2)^2 + 4^2}} = \frac{3}{\sqrt{4 + 16}} = \frac{3}{\sqrt{20}}$.

(A) The given equation is $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$.
For option $(A)$,if the slopes of the two lines are $m_1$ and $m_2$,then $m_1+m_2 = -\frac{2h}{b}$ and $m_1 m_2 = \frac{a}{b}$. If $m_1 = -m_2$,then $m_1+m_2 = 0$,which implies $-\frac{2h}{b} = 0$,so $h = 0$. Thus,option $(A)$ is correct.
For option $(B)$,if the lines are parallel,then $h^2 = ab$ and $bg^2 = af^2$. The condition $2f(gh+af) = 0$ is not generally true for parallel lines.
For option $(C)$,if the lines intersect at the origin,the constant term $c$ must be $0$ and the linear terms $2gx$ and $2fy$ must be $0$,so $g=f=c=0$. The condition $h^2=ab$ is for the lines to be parallel,not necessarily for intersection at the origin.
For option $(D)$,the $x$-coordinate of the point of intersection is given by $\frac{bg-hf}{h^2-ab}$. The condition $hf-bg > 0$ is equivalent to $bg-hf < 0$,which does not guarantee the $x$-coordinate is positive.

(D) For the circle $x^2+y^2+2kx+4y-4=0$,the centre $C_1 = (-k, -2)$ and radius $R_1 = \sqrt{k^2+4+4} = \sqrt{k^2+8}$.
For the circle $x^2+y^2+6x-2y+6=0$,the centre $C_2 = (-3, 1)$ and radius $R_2 = \sqrt{9+1-6} = 2$.
Since the circles touch each other,the distance between their centres $C_1C_2 = R_1 + R_2$ (assuming external touch).
$C_1C_2 = \sqrt{(-3+k)^2 + (1+2)^2} = \sqrt{k^2-6k+9+9} = \sqrt{k^2-6k+18}$.
Setting $C_1C_2 = R_1 + R_2$: $\sqrt{k^2-6k+18} = \sqrt{k^2+8} + 2$.
Squaring both sides: $k^2-6k+18 = k^2+8 + 4 + 4\sqrt{k^2+8}$.
$6-6k = 4\sqrt{k^2+8} \Rightarrow 3(1-k) = 2\sqrt{k^2+8}$.
Squaring again: $9(1-2k+k^2) = 4(k^2+8) \Rightarrow 9k^2-18k+9 = 4k^2+32$.
$5k^2-18k-23 = 0 \Rightarrow (k+1)(5k-23) = 0$.
So,$k = -1$ or $k = \frac{23}{5}$.
The centre is $(-k, -2)$. For the $4^{\text{th}}$ quadrant,the $x$-coordinate must be positive,so $-k > 0$,which means $k < 0$.
Thus,$k = -1$.

(D) For a point $(x_1, y_1)$ to not lie outside a circle $S(x, y) = 0$,it must lie inside or on the circle,i.e.,$S(x_1, y_1) \leq 0$.
For the circle $x^2+y^2-13=0$:
$(2)^2 + a^2 - 13 \leq 0$
$4 + a^2 - 13 \leq 0$
$a^2 - 9 \leq 0$
$(a+3)(a-3) \leq 0 \Rightarrow a \in [-3, 3] \quad (i)$
For the circle $x^2+y^2+x-2y-14=0$:
$(2)^2 + a^2 + 2 - 2a - 14 \leq 0$
$4 + a^2 + 2 - 2a - 14 \leq 0$
$a^2 - 2a - 8 \leq 0$
$(a-4)(a+2) \leq 0 \Rightarrow a \in [-2, 4] \quad (ii)$
Taking the intersection of $(i)$ and $(ii)$:
$a \in [-3, 3] \cap [-2, 4] = [-2, 3]$.

(B) The given system of circles is $x^2+y^2+2fy+\lambda(x^2+y^2+2gx+k)=0$.
Rearranging the terms,we get $(1+\lambda)x^2+(1+\lambda)y^2+2g\lambda x+2fy+\lambda k=0$.
Dividing by $(1+\lambda)$,we have $x^2+y^2+2\left(\frac{g\lambda}{1+\lambda}\right)x+2\left(\frac{f}{1+\lambda}\right)y+\frac{\lambda k}{1+\lambda}=0$.
For a point circle,the radius $r=0$,so $g'^2+f'^2-c'=0$.
Substituting the values,we get $\left(\frac{g\lambda}{1+\lambda}\right)^2+\left(\frac{f}{1+\lambda}\right)^2-\frac{\lambda k}{1+\lambda}=0$.
Multiplying by $(1+\lambda)^2$,we get $g^2\lambda^2+f^2-\lambda k(1+\lambda)=0$,which simplifies to $(g^2-k)\lambda^2-k\lambda+f^2=0$.
Let the roots be $\lambda_1$ and $\lambda_2$. The centers of the point circles are $A\left(\frac{-g\lambda_1}{1+\lambda_1}, \frac{-f}{1+\lambda_1}\right)$ and $B\left(\frac{-g\lambda_2}{1+\lambda_2}, \frac{-f}{1+\lambda_2}\right)$.
Since $\angle AOB = \frac{\pi}{2}$,the product of slopes $m_1 m_2 = -1$.
$m_1 = \frac{-f/(1+\lambda_1)}{-g\lambda_1/(1+\lambda_1)} = \frac{f}{g\lambda_1}$.
Thus,$\left(\frac{f}{g\lambda_1}\right)\left(\frac{f}{g\lambda_2}\right) = -1$ $\Rightarrow \frac{f^2}{g^2\lambda_1\lambda_2} = -1$.
From the quadratic equation,$\lambda_1\lambda_2 = \frac{f^2}{g^2-k}$.
Substituting this,$\frac{f^2}{g^2(f^2/(g^2-k))} = -1 \Rightarrow \frac{g^2-k}{g^2} = -1$.
$g^2-k = -g^2$ $\Rightarrow 2g^2 = k$ $\Rightarrow g^2 = \frac{k}{2}$.

(C) The given equations of the circles are:
$C_1: x^2+y^2+4x-6y-3=0$
$C_2: x^2+y^2+4x-2y+1=0$
Comparing with the general equation $x^2+y^2+2gx+2fy+c=0$:
For $C_1$: $g=2, f=-3, c=-3$. Centre $O_1 = (-2, 3)$,Radius $r_1 = \sqrt{g^2+f^2-c} = \sqrt{4+9+3} = \sqrt{16} = 4$.
For $C_2$: $g=2, f=-1, c=1$. Centre $O_2 = (-2, 1)$,Radius $r_2 = \sqrt{g^2+f^2-c} = \sqrt{4+1-1} = \sqrt{4} = 2$.
The distance between the centres $O_1O_2 = \sqrt{(-2 - (-2))^2 + (3-1)^2} = \sqrt{0^2 + 2^2} = 2$.
We observe that $|r_1 - r_2| = |4 - 2| = 2$.
Since the distance between the centres $O_1O_2 = |r_1 - r_2|$,the two circles touch each other internally.
When two circles touch internally,there is exactly $1$ common tangent.

(A) The family of circles passing through the intersection of the circle $S: x^2 + y^2 - a^2 = 0$ and the line $L: x \cos \alpha + y \sin \alpha - P = 0$ is given by $S + \lambda L = 0$.
$x^2 + y^2 - a^2 + \lambda(x \cos \alpha + y \sin \alpha - P) = 0$
$x^2 + y^2 + \lambda x \cos \alpha + \lambda y \sin \alpha - a^2 - \lambda P = 0$ $(i)$
The centre of this circle is $\left(-\frac{\lambda \cos \alpha}{2}, -\frac{\lambda \sin \alpha}{2}\right)$.
Since the line $x \cos \alpha + y \sin \alpha = P$ is the diameter $\overline{AB}$,the centre of the circle must lie on this line.
Substituting the centre into the line equation:
$\left(-\frac{\lambda \cos \alpha}{2}\right) \cos \alpha + \left(-\frac{\lambda \sin \alpha}{2}\right) \sin \alpha = P$
$-\frac{\lambda}{2} (\cos^2 \alpha + \sin^2 \alpha) = P$
$-\frac{\lambda}{2} = P \Rightarrow \lambda = -2P$
Substituting $\lambda = -2P$ into equation $(i)$:
$x^2 + y^2 - 2Px \cos \alpha - 2Py \sin \alpha - a^2 - (-2P)P = 0$
$x^2 + y^2 - 2Px \cos \alpha - 2Py \sin \alpha + 2P^2 - a^2 = 0$

(A) For the equation $x^2+2x-3=0$,the roots are $\alpha$ and $\beta$. By Vieta's formulas,$\alpha+\beta = -2$ and $\alpha\beta = -3$.
For the equation $y^2-y-6=0$,the roots are $\gamma$ and $\delta$. By Vieta's formulas,$\gamma+\delta = 1$ and $\gamma\delta = -6$.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
Expanding this,we get $x^2 - (x_1+x_2)x + x_1x_2 + y^2 - (y_1+y_2)y + y_1y_2 = 0$.
Substituting the values,we get $x^2 - (-2)x + (-3) + y^2 - (1)y + (-6) = 0$.
This simplifies to $x^2 + y^2 + 2x - y - 9 = 0$.

(A) The equation of the circle is $x^2+y^2-2x+4y-2=0$.
Given that the points $P(2,3)$ and $Q(K,-2)$ are conjugate with respect to the circle,the polar of point $P$ must pass through point $Q$.
The equation of the polar of a point $(x_1, y_1)$ with respect to the circle $x^2+y^2+2gx+2fy+c=0$ is given by $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$.
Substituting the values $x_1=2, y_1=3, g=-1, f=2, c=-2$,we get:
$x(2)+y(3)-1(x+2)+2(y+3)-2=0$
$2x+3y-x-2+2y+6-2=0$
$x+5y+2=0$.
Since the polar passes through $Q(K,-2)$,we substitute $x=K$ and $y=-2$ into the polar equation:
$K+5(-2)+2=0$
$K-10+2=0$
$K-8=0$
$K=8$.

(A) The equation of the polar of the point $(1,1)$ with respect to the circle $x^2+y^2-4x-6y+12=0$ is given by $x(1) + y(1) - 2(x+1) - 3(y+1) + 12 = 0$.
Simplifying this,we get $x + y - 2x - 2 - 3y - 3 + 12 = 0$,which results in $-x - 2y + 7 = 0$,or $x + 2y - 7 = 0$.
The inverse point $(h, k)$ is the foot of the perpendicular from the point $(1,1)$ to the polar line $x + 2y - 7 = 0$.
Using the formula for the foot of the perpendicular $\frac{h-x_1}{a} = \frac{k-y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$:
$\frac{h-1}{1} = \frac{k-1}{2} = -\frac{1(1) + 2(1) - 7}{1^2 + 2^2} = -\frac{1 + 2 - 7}{5} = -\frac{-4}{5} = \frac{4}{5}$.
From $\frac{h-1}{1} = \frac{4}{5}$,we get $h = 1 + \frac{4}{5} = \frac{9}{5}$.
From $\frac{k-1}{2} = \frac{4}{5}$,we get $k - 1 = \frac{8}{5}$,so $k = 1 + \frac{8}{5} = \frac{13}{5}$.
Therefore,$h + k = \frac{9}{5} + \frac{13}{5} = \frac{22}{5}$.

(D) The given equation is $9x^2-24xy+16y^2+\alpha x+\beta y+6=0$. This can be written as $(3x-4y)^2+\alpha x+\beta y+6=0$.
Since it represents parallel lines,let the lines be $(3x-4y+k_1)=0$ and $(3x-4y+k_2)=0$.
Their product is $(3x-4y+k_1)(3x-4y+k_2) = 9x^2-24xy+16y^2+3(k_1+k_2)x-4(k_1+k_2)y+k_1k_2=0$.
Comparing coefficients: $\alpha = 3(k_1+k_2)$,$\beta = -4(k_1+k_2)$,and $k_1k_2 = 6$.
Thus,$\frac{\alpha}{\beta} = \frac{3(k_1+k_2)}{-4(k_1+k_2)} = -\frac{3}{4}$.
The distance between parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ is $\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$.
Here,$\frac{|k_1-k_2|}{\sqrt{3^2+(-4)^2}} = 1 \implies |k_1-k_2| = 5$.
Since $(k_1-k_2)^2 = (k_1+k_2)^2 - 4k_1k_2$,we have $25 = (k_1+k_2)^2 - 24$,so $(k_1+k_2)^2 = 49$,meaning $k_1+k_2 = \pm 7$.
If one line passes through $(1,1)$,then $3(1)-4(1)+k = 0 \implies k = 1$.
If $k_1=1$,then $k_2 = 6$ (since $k_1k_2=6$),so $k_1+k_2 = 7$.
Then $\alpha = 3(7) = 21$ and $\beta = -4(7) = -28$.
Thus,$\frac{\alpha}{\beta} = \frac{21}{-28} = -\frac{3}{4}$.

(B) The given equations of the parallel tangents are $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$.
To make the coefficients of $x$ and $y$ identical,multiply the first equation by $2$:
$6x - 8y + 8 = 0$ and $6x - 8y - 7 = 0$.
The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
Here,the distance between the tangents is the diameter of the circle:
$d = \frac{|8 - (-7)|}{\sqrt{6^2 + (-8)^2}} = \frac{15}{\sqrt{36 + 64}} = \frac{15}{10} = \frac{3}{2}$.
Since the diameter is $\frac{3}{2}$,the radius $r = \frac{d}{2} = \frac{3}{4}$.
The area of the circle is $\pi r^2 = \pi \left(\frac{3}{4}\right)^2 = \frac{9\pi}{16}$ square units.

(B) The given equation of the circle is $x^2 + y^2 = 16$,which is of the form $x^2 + y^2 = a^2$ with $a^2 = 16$.
We know that the line $y = mx + C$ is a tangent to the circle $x^2 + y^2 = a^2$ if and only if $C^2 = a^2(1 + m^2)$.
Substituting $a^2 = 16$ into the condition,we get $C^2 = 16(1 + m^2)$.
Dividing by $16$,we have $\frac{C^2}{16} = 1 + m^2$.
Rearranging for $m^2$,we get $m^2 = \frac{C^2}{16} - 1 = \frac{C^2 - 16}{16}$.
Taking the square root on both sides,we get $m = \pm \frac{1}{4} \sqrt{C^2 - 16}$.
Thus,the correct option is $B$.

(A) Using the family of circles,the equation of the circle passing through the points of intersection of the circles $S_1: x^2+y^2+13x-3y=0$ and $S_2: 2x^2+2y^2+4x-7y-25=0$ is given by $S_2 + \lambda S_1 = 0$.
Substituting the equations,we get:
$2x^2+2y^2+4x-7y-25 + \lambda(x^2+y^2+13x-3y) = 0$.
Since the circle passes through $(1,1)$,we substitute $x=1$ and $y=1$ into the equation:
$2(1)^2+2(1)^2+4(1)-7(1)-25 + \lambda(1^2+1^2+13(1)-3(1)) = 0$.
$2+2+4-7-25 + \lambda(1+1+13-3) = 0$.
$-24 + \lambda(12) = 0$.
$12\lambda = 24 \Rightarrow \lambda = 2$.
Substituting $\lambda = 2$ back into the family equation:
$2x^2+2y^2+4x-7y-25 + 2(x^2+y^2+13x-3y) = 0$.
$2x^2+2y^2+4x-7y-25 + 2x^2+2y^2+26x-6y = 0$.
$4x^2+4y^2+30x-13y-25 = 0$.

(C) The given circles are:
$x^2+y^2-4x-6y+12=0 \dots(1)$
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get center $C_1 = (2, 3)$ and radius $r_1 = \sqrt{2^2+3^2-12} = \sqrt{13-12} = 1$.
$x^2+y^2+4x-2y-4=0 \dots(2)$
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get center $C_2 = (-2, 1)$ and radius $r_2 = \sqrt{(-2)^2+1^2-(-4)} = \sqrt{4+1+4} = 3$.
The internal centre of similitude divides the line segment joining the centers $C_1$ and $C_2$ internally in the ratio $r_1 : r_2 = 1 : 3$.
Using the section formula,the coordinates are:
$C = \left(\frac{r_1 x_2 + r_2 x_1}{r_1+r_2}, \frac{r_1 y_2 + r_2 y_1}{r_1+r_2}\right)$
$C = \left(\frac{1(-2) + 3(2)}{1+3}, \frac{1(1) + 3(3)}{1+3}\right)$
$C = \left(\frac{-2+6}{4}, \frac{1+9}{4}\right) = \left(\frac{4}{4}, \frac{10}{4}\right) = \left(1, \frac{5}{2}\right)$.

(B) The given circles are $S_1: x^2+y^2-2x-9=0$ and $S_2: x^2+y^2-4y-1=0$.
The centre $C_1$ and radius $r_1$ of $S_1$ are $(1, 0)$ and $\sqrt{1^2+0^2-(-9)} = \sqrt{10}$.
The centre $C_2$ and radius $r_2$ of $S_2$ are $(0, 2)$ and $\sqrt{0^2+2^2-(-1)} = \sqrt{5}$.
The distance $d$ between the centres $C_1(1, 0)$ and $C_2(0, 2)$ is $d = \sqrt{(1-0)^2 + (0-2)^2} = \sqrt{1+4} = \sqrt{5}$.
The angle $\theta$ between the circles is given by $\cos \theta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2}$.
Substituting the values: $\cos \theta = \frac{10 + 5 - 5}{2 \times \sqrt{10} \times \sqrt{5}} = \frac{10}{2 \times \sqrt{50}} = \frac{10}{2 \times 5\sqrt{2}} = \frac{10}{10\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Therefore,$\theta = \frac{\pi}{4}$.

(C) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$.
Since it passes through $(a, b)$,we have $a^2+b^2+2ga+2fb+c=0$,which implies $c = -a^2-b^2-2ga-2fb$.
Since the circle cuts $x^2+y^2-2x+4y-4=0$ orthogonally,the condition $2g_1g_2 + 2f_1f_2 = c_1+c_2$ gives:
$2g(-1) + 2f(2) = c - 4$.
Substituting $c$:
$-2g + 4f = -a^2-b^2-2ga-2fb - 4$.
Rearranging terms:
$g(2a-2) + f(2b+4) = -a^2-b^2-4$.
Let the centre be $(x, y) = (-g, -f)$,so $g = -x$ and $f = -y$.
Substituting these:
$-x(2a-2) - y(2b+4) = -a^2-b^2-4$.
$x(2a-2) + y(2b+4) = a^2+b^2+4$.
Dividing by $2$:
$(a-1)x + (b+2)y = \frac{a^2+b^2+4}{2}$.

(A) The equation of the chord of a circle $S \equiv x^2+y^2+2gx+2fy+c=0$ with a given midpoint $M(x_1, y_1)$ is given by $T=S_1$,which is $x x_1 + y y_1 + g(x+x_1) + f(y+y_1) + c = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$.
Given the circle $x^2+y^2-2x=0$,we have $g=-1, f=0, c=0$.
The chord passes through the point $(0,0)$. Substituting $(0,0)$ into the equation $T=S_1$:
$0(x_1) + 0(y_1) - 1(0+x_1) + 0(0+y_1) + 0 = x_1^2 + y_1^2 - 2x_1$.
$-x_1 = x_1^2 + y_1^2 - 2x_1$.
$x_1^2 + y_1^2 - x_1 = 0$.
Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $x^2+y^2-x=0$.

(A) For the circle $x^2+y^2=1$:
Centre $C_1 = (0,0)$,Radius $R_1 = 1$.
For the circle $x^2+y^2-2x-6y+6=0$:
Centre $C_2 = (1,3)$,Radius $R_2 = \sqrt{1^2+3^2-6} = \sqrt{4} = 2$.
The distance between the centres $C_1$ and $C_2$ is $d = \sqrt{(1-0)^2 + (3-0)^2} = \sqrt{1+9} = \sqrt{10}$.
Since $\sqrt{10} \approx 3.16$ and $R_1+R_2 = 1+2 = 3$,we have $d > R_1+R_2$.
Because the distance between the centres is greater than the sum of the radii,the two circles do not intersect and lie outside each other.
Therefore,there are $4$ common tangents (two direct and two transverse).

(D) Given circles are $S_1 \equiv x^2+y^2-2x+18y+78=0$ and $S_2 \equiv x^2+y^2+8x-6y-200=0$.
Centre $C_1 = (1, -9)$ and radius $r_1 = \sqrt{1^2+(-9)^2-78} = \sqrt{1+81-78} = 2$.
Centre $C_2 = (-4, 3)$ and radius $r_2 = \sqrt{(-4)^2+3^2-(-200)} = \sqrt{16+9+200} = 15$.
Distance $C_1C_2 = \sqrt{(-4-1)^2+(3-(-9))^2} = \sqrt{(-5)^2+12^2} = \sqrt{25+144} = 13$.
Since $r_2 - r_1 = 15 - 2 = 13 = C_1C_2$,the circles touch each other internally.
The equation of the common tangent is $S_1 - S_2 = 0$.
$(x^2+y^2-2x+18y+78) - (x^2+y^2+8x-6y-200) = 0$.
$-10x + 24y + 278 = 0$,which simplifies to $-5x + 12y + 139 = 0$.
Checking option $D$: $-5\left(\frac{-2}{5}\right) + 12\left(\frac{-47}{4}\right) + 139 = 2 - 141 + 139 = 0$.
Thus,the point $\left(\frac{-2}{5}, \frac{-47}{4}\right)$ lies on the common tangent.

(A) The given equations of the circles are $x^2+y^2-2x-2y-23=0 \dots (1)$ and $x^2+y^2-4x-4y-1=0 \dots (2)$.
For circle $(1)$,the centre $C_1 = (1, 1)$ and radius $r_1 = \sqrt{1^2+1^2-(-23)} = \sqrt{25} = 5$.
For circle $(2)$,the centre $C_2 = (2, 2)$ and radius $r_2 = \sqrt{2^2+2^2-(-1)} = \sqrt{9} = 3$.
The distance between the centres $C_1$ and $C_2$ is $d = \sqrt{(2-1)^2+(2-1)^2} = \sqrt{1+1} = \sqrt{2} \approx 1.414$.
We compare $d$ with the difference of the radii $|r_1 - r_2| = |5 - 3| = 2$.
Since $d < |r_1 - r_2|$ (as $\sqrt{2} < 2$),the smaller circle lies completely inside the larger circle.
Therefore,the number of common tangents that can be drawn is $0$.

(D) The radical axis of two circles is the locus of points from which the lengths of the tangents to the two circles are equal.
When two circles touch each other,their radical axis is the common tangent at the point of contact.
Since the circles touch at $(0,0)$,their radical axis is the common tangent at $(0,0)$.

(A) The given circles are $x^2+y^2+2hx+2ky=0$ and $x^2+y^2+2px+2qy=0$.
Both circles pass through the origin $(0,0)$.
Two circles $x^2+y^2+2g_1x+2f_1y=0$ and $x^2+y^2+2g_2x+2f_2y=0$ touch each other at the origin if their centers are collinear with the origin,which implies $\frac{g_1}{f_1} = \frac{g_2}{f_2}$,or $g_1f_2 = g_2f_1$.
Here,$g_1=h, f_1=k, g_2=p, f_2=q$.
Thus,the condition for touching at the origin is $hq = pk$,which implies $hq - pk = 0$.
Also,since $hq = pk$ and $p, k \neq 0$,we have $\frac{hq}{pk} = 1$.
Substituting these values into the expression:
$hq - pk - \frac{hq}{pk} = 0 - 1 = -1$.

(B) The equation of the common chord of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$.
Given circles are $S_1: x^2+y^2+2x+3y+1=0$ and $S_2: x^2+y^2+4x+3y+2=0$.
Subtracting $S_2$ from $S_1$,we get $(x^2+y^2+2x+3y+1) - (x^2+y^2+4x+3y+2) = 0$,which simplifies to $-2x - 1 = 0$,or $x = -\frac{1}{2}$.
Substitute $x = -\frac{1}{2}$ into the first circle equation: $(-\frac{1}{2})^2 + y^2 + 2(-\frac{1}{2}) + 3y + 1 = 0$.
$\frac{1}{4} + y^2 - 1 + 3y + 1 = 0 \Rightarrow y^2 + 3y + \frac{1}{4} = 0$.
Multiplying by $4$,we get $4y^2 + 12y + 1 = 0$.
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $y = \frac{-12 \pm \sqrt{144 - 16}}{8} = \frac{-12 \pm \sqrt{128}}{8} = \frac{-12 \pm 8\sqrt{2}}{8} = -\frac{3}{2} \pm \sqrt{2}$.
The endpoints of the common chord are $(-\frac{1}{2}, -\frac{3}{2} + \sqrt{2})$ and $(-\frac{1}{2}, -\frac{3}{2} - \sqrt{2})$.
The length of the common chord is the distance between these points: $\sqrt{(-\frac{1}{2} - (-\frac{1}{2}))^2 + ((-\frac{3}{2} + \sqrt{2}) - (-\frac{3}{2} - \sqrt{2}))^2} = \sqrt{0^2 + (2\sqrt{2})^2} = 2\sqrt{2}$ units.

(A) The first circle is $x^2+y^2-2x-6y+(10-r^2)=0$. Its center $C_1 = (1, 3)$ and radius $r_1 = \sqrt{1^2+3^2-(10-r^2)} = \sqrt{1+9-10+r^2} = |r|$.
The second circle is $x^2+y^2-8x+2y+8=0$. Its center $C_2 = (4, -1)$ and radius $r_2 = \sqrt{4^2+(-1)^2-8} = \sqrt{16+1-8} = 3$.
The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(4-1)^2+(-1-3)^2} = \sqrt{3^2+(-4)^2} = \sqrt{9+16} = 5$.
For two circles to have a common chord of non-zero length,they must intersect at two distinct points. This occurs if $|r_1 - r_2| < d < r_1 + r_2$.
Substituting the values: $| |r| - 3 | < 5 < |r| + 3$.
Case $1$: $|r| + 3 > 5 \implies |r| > 2$.
Case $2$: $| |r| - 3 | < 5 \implies -5 < |r| - 3 < 5 \implies -2 < |r| < 8$.
Combining these,we get $2 < |r| < 8$.

(B) For the first circle $(x-2)^2+(y-3)^2=5^2$,the center $C_1 = (2, 3)$ and radius $r_1 = 5$.
For the second circle $25x^2+25y^2-40x-70y-160=0$,dividing by $25$ gives $x^2+y^2-\frac{8}{5}x-\frac{14}{5}y-\frac{32}{5}=0$.
Completing the square: $(x-\frac{4}{5})^2+(y-\frac{7}{5})^2 = \frac{32}{5} + \frac{16}{25} + \frac{49}{25} = \frac{160+16+49}{25} = \frac{225}{25} = 9 = 3^2$.
Thus,the center $C_2 = (\frac{4}{5}, \frac{7}{5})$ and radius $r_2 = 3$.
Since the circles touch internally,the point of contact $(\alpha, \beta)$ divides the line segment joining the centers $C_1$ and $C_2$ externally in the ratio $r_1 : r_2$.
$(\alpha, \beta) = \left(\frac{r_1 x_2 - r_2 x_1}{r_1 - r_2}, \frac{r_1 y_2 - r_2 y_1}{r_1 - r_2}\right) = \left(\frac{5(\frac{4}{5}) - 3(2)}{5-3}, \frac{5(\frac{7}{5}) - 3(3)}{5-3}\right) = \left(\frac{4-6}{2}, \frac{7-9}{2}\right) = (-1, -1)$.
Therefore,$\alpha + \beta = -1 + (-1) = -2$.

(B) The family of circles passing through the points of intersection of two circles $S_1: x^2+y^2+2x-2y-2=0$ and $S_2: x^2+y^2-2x+2y-7=0$ is given by $S_1 + \lambda S_2 = 0$.
$(x^2+y^2+2x-2y-2) + \lambda(x^2+y^2-2x+2y-7) = 0$
$(1+\lambda)x^2 + (1+\lambda)y^2 + 2(1-\lambda)x + 2(\lambda-1)y - (2+7\lambda) = 0$
Dividing by $(1+\lambda)$,we get the equation of the circle:
$x^2 + y^2 + 2\left(\frac{1-\lambda}{1+\lambda}\right)x + 2\left(\frac{\lambda-1}{1+\lambda}\right)y - \frac{2+7\lambda}{1+\lambda} = 0$
The centre of this circle is $\left(-\frac{1-\lambda}{1+\lambda}, -\frac{\lambda-1}{1+\lambda}\right) = \left(\frac{\lambda-1}{\lambda+1}, \frac{1-\lambda}{\lambda+1}\right)$.
Since the centre lies on the line $x-y-1=0$,we substitute the coordinates:
$\frac{\lambda-1}{\lambda+1} - \frac{1-\lambda}{\lambda+1} - 1 = 0$
$\frac{\lambda-1 - 1 + \lambda - \lambda - 1}{\lambda+1} = 0$
$\lambda - 3 = 0 \Rightarrow \lambda = 3$.
Substituting $\lambda = 3$ into the centre coordinates:
$x = \frac{3-1}{3+1} = \frac{2}{4} = \frac{1}{2}$
$y = \frac{1-3}{3+1} = \frac{-2}{4} = -\frac{1}{2}$
Thus,the centre is $\left(\frac{1}{2}, -\frac{1}{2}\right)$.

(A) The family of circles passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + \lambda S_2 = 0$.
Substituting the given equations:
$(x^2+y^2+2x+3y+1) + \lambda(x^2+y^2+4x+3y+2) = 0$.
Since the circle passes through the point $(-1, 1)$,we substitute $x = -1$ and $y = 1$:
$((-1)^2 + (1)^2 + 2(-1) + 3(1) + 1) + \lambda((-1)^2 + (1)^2 + 4(-1) + 3(1) + 2) = 0$.
$(1 + 1 - 2 + 3 + 1) + \lambda(1 + 1 - 4 + 3 + 2) = 0$.
$4 + 3\lambda = 0 \Rightarrow \lambda = -\frac{4}{3}$.
Substituting $\lambda = -\frac{4}{3}$ back into the family equation:
$(x^2+y^2+2x+3y+1) - \frac{4}{3}(x^2+y^2+4x+3y+2) = 0$.
$3(x^2+y^2+2x+3y+1) - 4(x^2+y^2+4x+3y+2) = 0$.
$3x^2+3y^2+6x+9y+3 - 4x^2-4y^2-16x-12y-8 = 0$.
$-x^2-y^2-10x-3y-5 = 0$.
$x^2+y^2+10x+3y+5 = 0$.

(D) The polar of the circle $x^2+y^2+2gx+2fy+c=0$ with respect to the pole $(x_1, y_1)$ is given by $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$.
For $S_1 \equiv x^2+y^2+6y+7=0$ with pole $(-1, 2)$,the polar is:
$-x+2y+3(y+2)+7=0 \Rightarrow -x+5y+13=0$ (Equation $1$).
For $S_2 \equiv x^2+y^2+6x+1=0$ with pole $(-1, 2)$,the polar is:
$-x+2y+3(x-1)+1=0$ $\Rightarrow 2x+2y-2=0$ $\Rightarrow x+y-1=0$ (Equation $2$).
To check the intersection,we solve the system:
$-x+5y+13=0$
$x+y-1=0$
Adding these equations gives $6y+12=0$,so $y=-2$.
Substituting $y=-2$ into $x+y-1=0$ gives $x-2-1=0$,so $x=3$.
The polars intersect at the point $(3, -2)$,which is a non-zero point.

(A) The coordinate of $P$ is $\left(5 \cos \frac{2 \pi}{3}, 5 \sin \frac{2 \pi}{3}\right) = \left(-\frac{5}{2}, \frac{5 \sqrt{3}}{2}\right)$.
The slope of the tangent at $\theta = \frac{2 \pi}{3}$ is $m = -\cot \left(\frac{2 \pi}{3}\right) = \frac{1}{\sqrt{3}}$.
The equation of the tangent line is $y - \frac{5 \sqrt{3}}{2} = \frac{1}{\sqrt{3}}\left(x + \frac{5}{2}\right)$,which simplifies to $x - \sqrt{3}y + 10 = 0$.
To find the intersection point $Q$ with the line $y = 6$,we substitute $y = 6$ into the tangent equation: $x - 6\sqrt{3} + 10 = 0 \Rightarrow x = 6\sqrt{3} - 10$. Thus,$Q = (6\sqrt{3} - 10, 6)$.
To find the intersection point $R$ with the line $x = -6$,we substitute $x = -6$ into the tangent equation: $-6 - \sqrt{3}y + 10 = 0$ $\Rightarrow \sqrt{3}y = 4$ $\Rightarrow y = \frac{4}{\sqrt{3}}$. Thus,$R = (-6, \frac{4}{\sqrt{3}})$.
The third vertex of the triangle is $S = (-6, 6)$.
The area of $\triangle RSQ$ is given by $\frac{1}{2} |x_R(y_S - y_Q) + x_S(y_Q - y_R) + x_Q(y_R - y_S)|$.
Area $= \frac{1}{2} |(-6)(6 - 6) + (-6)(6 - \frac{4}{\sqrt{3}}) + (6\sqrt{3} - 10)(\frac{4}{\sqrt{3}} - 6)|$.
Area $= \frac{1}{2} |0 - 36 + \frac{24}{\sqrt{3}} + (24 - 36\sqrt{3} - \frac{40}{\sqrt{3}} + 60)| = \frac{1}{2} |48 - 36\sqrt{3} - \frac{16}{\sqrt{3}}| = |24 - 18\sqrt{3} - \frac{8}{\sqrt{3}}| = |\frac{24\sqrt{3} - 54 - 8}{\sqrt{3}}| = \frac{62 - 24\sqrt{3}}{\sqrt{3}}$.

(B) The given circles are:
$S_1 \equiv x^2+y^2-8x-2y+8=0$
$S_2 \equiv x^2+y^2+6x+8y-24=0$
$S_3 \equiv x^2+y^2-2x+2y+2=0$
The radical axis of $S_1$ and $S_2$ is given by $S_1 - S_2 = 0$:
$(x^2+y^2-8x-2y+8) - (x^2+y^2+6x+8y-24) = 0$
$-14x - 10y + 32 = 0 \Rightarrow 7x + 5y = 16 \quad \dots(1)$
The radical axis of $S_2$ and $S_3$ is given by $S_2 - S_3 = 0$:
$(x^2+y^2+6x+8y-24) - (x^2+y^2-2x+2y+2) = 0$
$8x + 6y - 26 = 0 \Rightarrow 4x + 3y = 13 \quad \dots(2)$
To find the radical centre $(a, b)$,we solve the system of equations $(1)$ and $(2)$:
From $(2)$,$y = \frac{13-4x}{3}$. Substituting this into $(1)$:
$7x + 5(\frac{13-4x}{3}) = 16$
$21x + 65 - 20x = 48$
$x = -17$
Substituting $x = -17$ into the expression for $y$:
$y = \frac{13 - 4(-17)}{3} = \frac{13 + 68}{3} = \frac{81}{3} = 27$
Thus,the radical centre $(a, b) = (-17, 27)$.
Therefore,$a+b = -17 + 27 = 10$.

(B) The vertices of the triangle are the intersection points of the lines:
$x+y=6$ $(1)$,$2x+y=4$ $(2)$,and $x+2y=5$ $(3)$.
Solving $(1)$ and $(2)$: $x = -2, y = 8$. Vertex $A = (-2, 8)$.
Solving $(2)$ and $(3)$: $x = 1, y = 2$. Vertex $B = (1, 2)$.
Solving $(1)$ and $(3)$: $x = 7, y = -1$. Vertex $C = (7, -1)$.
Let the centre of the circle be $(a, b)$. The distance from the centre to each vertex is equal (radius $R$).
$(a+2)^2 + (b-8)^2 = (a-1)^2 + (b-2)^2 = (a-7)^2 + (b+1)^2$.
From $(a+2)^2 + (b-8)^2 = (a-1)^2 + (b-2)^2$:
$a^2 + 4a + 4 + b^2 - 16b + 64 = a^2 - 2a + 1 + b^2 - 4b + 4$.
$6a - 12b = -63 \implies 2a - 4b = -21$.
From $(a-1)^2 + (b-2)^2 = (a-7)^2 + (b+1)^2$:
$a^2 - 2a + 1 + b^2 - 4b + 4 = a^2 - 14a + 49 + b^2 + 2b + 1$.
$12a - 6b = 45 \implies 4a - 2b = 15$.
Solving the system $2a - 4b = -21$ and $4a - 2b = 15$:
Multiply the first by $2$: $4a - 8b = -42$.
Subtracting: $(4a - 2b) - (4a - 8b) = 15 - (-42) \implies 6b = 57 \implies b = \frac{57}{6} = \frac{19}{2}$.
Substituting $b = \frac{19}{2}$ into $4a - 2b = 15$:
$4a - 19 = 15 \implies 4a = 34 \implies a = \frac{34}{4} = \frac{17}{2}$.
Thus,the centre $(a, b)$ is $(\frac{17}{2}, \frac{19}{2})$.

(C) Let the centre of the circle be $(h, k)$.
Since the circle passes through the origin $(0, 0)$,its radius $R$ is given by $R^2 = h^2 + k^2$.
The equation of the circle is $(x-h)^2 + (y-k)^2 = h^2 + k^2$,which simplifies to $x^2 + y^2 - 2hx - 2ky = 0$.
The circle cuts a chord of length $2$ units on the line $x=1$. The length of the chord is given by $2\sqrt{R^2 - d^2} = 2$,where $d$ is the perpendicular distance from the centre $(h, k)$ to the line $x=1$.
Thus,$\sqrt{R^2 - d^2} = 1$,or $R^2 - d^2 = 1$.
Here,$R^2 = h^2 + k^2$ and $d = |h-1|$.
Substituting these values,we get $(h^2 + k^2) - (h-1)^2 = 1$.
$h^2 + k^2 - (h^2 - 2h + 1) = 1$
$h^2 + k^2 - h^2 + 2h - 1 = 1$
$k^2 + 2h - 1 = 1$
$k^2 = 2 - 2h$
Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = 2(1-x)$,which represents a parabola.

(D) For a parabola,the distance from any point $P(x, y)$ on the parabola to the focus $S$ is equal to the perpendicular distance from $P$ to the directrix $L$.
$PS = PN$
$\Rightarrow \sqrt{(x-2)^2+(y+3)^2} = \left|\frac{3x-2y+5}{\sqrt{3^2+(-2)^2}}\right|$
$\Rightarrow (x-2)^2+(y+3)^2 = \frac{(3x-2y+5)^2}{13}$
$\Rightarrow 13(x^2-4x+4+y^2+6y+9) = 9x^2+4y^2+25-12xy-20y+30x$
$\Rightarrow 13x^2-52x+52+13y^2+78y+117 = 9x^2+4y^2-12xy+30x-20y+25$
$\Rightarrow 4x^2+12xy+9y^2-82x+98y+144 = 0$
Comparing this with the given equation $ax^2+2hxy+by^2-82x+98y+144=0$,we get $a=4, h=6, b=9$.
Now,the equation $ax^2+2hxy+by^2=0$ becomes $4x^2+12xy+9y^2=0$.
This can be written as $(2x+3y)^2=0$.
Since the discriminant $h^2-ab = 6^2-(4)(9) = 36-36=0$,it represents two coincident lines.

(B) The given equation of the parabola is $(2 x - 3 y - 5)^2 = 20(3 x + 2 y + 1)$.
We rewrite this in the form $\left( \frac{2 x - 3 y - 5}{\sqrt{2^2 + (-3)^2}} \right)^2 = \frac{20}{13} \left( \frac{3 x + 2 y + 1}{\sqrt{3^2 + 2^2}} \right) \sqrt{13}$.
Let $Y = \frac{2 x - 3 y - 5}{\sqrt{13}}$ and $X = \frac{3 x + 2 y + 1}{\sqrt{13}}$.
The equation becomes $Y^2 = \frac{20}{\sqrt{13}} X$.
Comparing with $Y^2 = 4 a X$,we get $4 a = \frac{20}{\sqrt{13}}$,so $a = \frac{5}{\sqrt{13}}$.
The directrix is given by $X = -a$,which is $\frac{3 x + 2 y + 1}{\sqrt{13}} = -\frac{5}{\sqrt{13}}$.
Thus,$3 x + 2 y + 1 = -5$,or $3 x + 2 y + 6 = 0$.

(D) Given equation: $y^2-4x-8y-12=0$
Rearranging terms: $y^2-8y = 4x+12$
Completing the square: $y^2-8y+16 = 4x+12+16$
$(y-4)^2 = 4x+28$
$(y-4)^2 = 4(x+7)$
Let $Y = y-4$ and $X = x+7$. Then the equation becomes $Y^2 = 4X$.
Comparing with the standard form $Y^2 = 4aX$,we get $4a = 4$,so $a = 1$.
The parametric equations for $Y^2 = 4aX$ are $X = at^2$ and $Y = 2at$.
Substituting $a = 1$: $X = t^2$ and $Y = 2t$.
Substituting back $X = x+7$ and $Y = y-4$:
$x+7 = t^2 \Rightarrow x = -7+t^2$
$y-4 = 2t \Rightarrow y = 4+2t$
Thus,the correct option is $D$.

(D) The general equation of a parabola opening to the right is $(y - k)^2 = 4a(x - h)$,where $(h, k)$ is the vertex.
Given vertex $(h, k) = (1, -2)$ and focus $(h + a, k) = \left(\frac{5}{4}, -2\right)$.
Calculating $a$: $h + a = \frac{5}{4}$ $\Rightarrow 1 + a = \frac{5}{4}$ $\Rightarrow a = \frac{1}{4}$.
Substituting the values into the equation:
$(y - (-2))^2 = 4 \times \frac{1}{4}(x - 1)$
$(y + 2)^2 = (x - 1)$.
Now,check the given options:
For option $(D)$,substitute $(10, 1)$ into the equation:
$(1 + 2)^2 = 3^2 = 9$
$(10 - 1) = 9$.
Since $9 = 9$,the point $(10, 1)$ lies on the parabola.

(C) Given parametric equations are $y=2+4t$ and $x=-2+2t^2$.
From the first equation,$t = \frac{y-2}{4}$.
Substituting this into the second equation: $x = -2 + 2\left(\frac{y-2}{4}\right)^2 = -2 + 2\frac{(y-2)^2}{16} = -2 + \frac{(y-2)^2}{8}$.
Rearranging gives $(y-2)^2 = 8(x+2)$.
Comparing this with $(y-k)^2 = 4a(x-h)$,we get $h=-2, k=2$,and $4a=8$,so $a=2$.
The ends of the latus rectum are at $(h+a, k \pm 2a)$,which are $(-2+2, 2 \pm 4)$,i.e.,$(0, 6)$ and $(0, -2)$.
For $y=6$,$2+4t=6 \implies 4t=4 \implies t=1$.
For $y=-2$,$2+4t=-2 \implies 4t=-4 \implies t=-1$.
Thus,$\alpha=1$ and $\beta=-1$.
Therefore,$\alpha \beta = (1)(-1) = -1$.

(B) The equation of a parabola with a horizontal axis is $x = ay^2 + by + c$.
Substituting the given points $(-2, 1)$,$(1, 2)$,and $(-1, 3)$:
$1) -2 = a + b + c$
$2) 1 = 4a + 2b + c$
$3) -1 = 9a + 3b + c$
Subtracting $(1)$ from $(2)$: $3a + b = 3$ $(4)$
Subtracting $(2)$ from $(3)$: $5a + b = -2$ $(5)$
Subtracting $(4)$ from $(5)$: $2a = -5 \Rightarrow a = -\frac{5}{2}$.
Substituting $a$ into $(4)$: $3(-\frac{5}{2}) + b = 3 \Rightarrow b = 3 + \frac{15}{2} = \frac{21}{2}$.
Substituting $a$ and $b$ into $(1)$: $-\frac{5}{2} + \frac{21}{2} + c = -2$ $\Rightarrow 8 + c = -2$ $\Rightarrow c = -10$.
The parabola is $x = -\frac{5}{2}y^2 + \frac{21}{2}y - 10$.
Completing the square for $y$: $x = -\frac{5}{2}(y^2 - \frac{21}{5}y) - 10 = -\frac{5}{2}(y - \frac{21}{10})^2 + \frac{5}{2}(\frac{441}{100}) - 10 = -\frac{5}{2}(y - \frac{21}{10})^2 + \frac{441}{40} - \frac{400}{40} = -\frac{5}{2}(y - \frac{21}{10})^2 + \frac{41}{40}$.
Rearranging: $(y - \frac{21}{10})^2 = -\frac{2}{5}(x - \frac{41}{40})$.
This is of the form $(y - k)^2 = 4A(x - h)$,where $k = \frac{21}{10}$.
The $y$-coordinate of the focus is $k = \frac{21}{10}$.

(A) Given equation of parabola is $x^2-8 x+12 y+15=0$.
Completing the square for $x$:
$x^2-8 x+16 = -12 y - 15 + 16$
$(x-4)^2 = -12 y + 1$
$(x-4)^2 = -12(y - \frac{1}{12})$.
Comparing this with the standard form $(x-h)^2 = -4a(y-k)$,we get $h=4$,$k=\frac{1}{12}$,and $4a=12$,so $a=3$.
The parametric equations for $(x-h)^2 = -4a(y-k)$ are $x = h + 2at$ and $y = k - at^2$.
Substituting the values,we get $x = 4 + 2(3)t = 4 + 6t$ and $y = \frac{1}{12} - 3t^2$.
Thus,the correct option is $A$.

(C) The given equation of the parabola is $x^2=16y$.
Comparing this with the standard form $x^2=4ay$,we get $4a=16$,which implies $a=4$.
The focus of the parabola is $F(0, a) = (0, 4)$.
The latus rectum is the line $y=4$.
Substituting $y=4$ in the parabola equation $x^2=16y$,we get $x^2=16(4)=64$,so $x=\pm 8$.
The endpoints of the latus rectum are $P(8, 4)$ and $Q(-8, 4)$.
The vertex of the parabola is $O(0, 0)$.
The triangle is formed by the vertices $O(0, 0)$,$P(8, 4)$,and $Q(-8, 4)$.
The base of the triangle $PQ$ has length $8 - (-8) = 16$ units.
The height of the triangle from the vertex $O$ to the line $PQ$ is the $y$-coordinate of the latus rectum,which is $4$ units.
Area of $\triangle OPQ = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 16 \times 4 = 32$ sq. units.

(D) . The equation is $y^2-2y+1 = -4x-3+1 \implies (y-1)^2 = -4(x+\frac{1}{2})$. Vertex is $\left(-\frac{1}{2}, 1\right)$,which is $(III)$.
$(B)$. The equation is $x^2+8x+16 = -12y-4+16 \implies (x+4)^2 = -12(y-1)$. Vertex is $(-4, 1)$,which is $(V)$.
$(C)$. The equation is $y^2-2y+1 = x-2+1 \implies (y-1)^2 = 1(x-1)$. Here $4a=1 \implies a=\frac{1}{4}$. Focus is $(h+a, k) = (1+\frac{1}{4}, 1) = \left(\frac{5}{4}, 1\right)$,which is $(I)$.
$(D)$. The equation is $x^2-2x+1 = 8y+23+1 \implies (x-1)^2 = 8(y+3)$. Here $4a=8 \implies a=2$. Focus is $(h, k+a) = (1, -3+2) = (1, -1)$,which is $(IV)$.
Thus,the correct match is $A-III, B-V, C-I, D-IV$.

(A) Let the equation of the parabola be $y = ax^2 + bx + c$ ... $(i)$
Since the parabola passes through $(0, 2)$,we have $2 = a(0)^2 + b(0) + c$,so $c = 2$.
Since it passes through $(1, -3)$,we have $-3 = a(1)^2 + b(1) + 2$,which simplifies to $a + b = -5$ ... (ii)
Since it passes through $(-1, 5)$,we have $5 = a(-1)^2 + b(-1) + 2$,which simplifies to $a - b = 3$ ... (iii)
Adding equations (ii) and (iii),we get $2a = -2$,so $a = -1$.
Substituting $a = -1$ into equation (ii),we get $-1 + b = -5$,so $b = -4$.
Thus,the equation of the parabola is $y = -x^2 - 4x + 2$.
Since $(2, k)$ lies on this parabola,we substitute $x = 2$ into the equation:
$k = -(2)^2 - 4(2) + 2 = -4 - 8 + 2 = -10$.

(D) The given equation is $25[(x-2)^2+(y-3)^2]=(3x-4y+7)^2$.
Dividing by $25$,we get $(x-2)^2+(y-3)^2 = \left(\frac{3x-4y+7}{5}\right)^2$.
This is in the form $SP^2 = PM^2$,where $S(2,3)$ is the focus and $3x-4y+7=0$ is the directrix.
Here,the distance from the focus to the directrix is $a = \left|\frac{3(2)-4(3)+7}{\sqrt{3^2+(-4)^2}}\right| = \left|\frac{6-12+7}{5}\right| = \frac{1}{5}$.
The length of the latus rectum of a parabola is $4a$.
Therefore,length $= 4 \times \frac{1}{5} = \frac{4}{5}$.

(C) Let $P$ be $(at^2, 2at)$ and $Q$ be $(h, k)$. Given $A = (-1, 3)$.
Since $Q$ divides $AP$ in the ratio $3:2$,by section formula:
$h = \frac{3(at^2) + 2(-1)}{3+2} = \frac{3at^2 - 2}{5} \implies 5h = 3at^2 - 2 \implies 3at^2 = 5h + 2 \quad \dots(i)$
$k = \frac{3(2at) + 2(3)}{3+2} = \frac{6at + 6}{5} \implies 5k = 6at + 6 \implies at = \frac{5k-6}{6} \quad \dots(ii)$
Squaring $(ii)$,we get $a^2t^2 = \frac{(5k-6)^2}{36}$.
From $(i)$,$t^2 = \frac{5h+2}{3a}$.
Substituting $t^2$ in the squared equation:
$a^2 \left(\frac{5h+2}{3a}\right) = \frac{(5k-6)^2}{36} \implies 12a(5h+2) = (5k-6)^2$.
This is of the form $(Y-k_0)^2 = 4A(X-h_0)$,where $Y = 5k-6$,$X = 5h+2$,and $4A = 12a \implies A = 3a$.
The focus of $Y^2 = 4AX$ is $(A, 0)$.
Thus,$5k-6 = 0 \implies k = 6/5$ and $5h+2 = 3a \implies h = \frac{3a-2}{5}$.
Wait,re-evaluating the point $A$ from the image: $A$ is $(-1, 3)$.
$h = \frac{3at^2 - 2}{5}$,$k = \frac{6at + 6}{5}$.
$5h+2 = 3at^2$,$5k-6 = 6at \implies (5k-6)^2 = 36a^2t^2 = 36a \left(\frac{5h+2}{3}\right) = 12a(5h+2)$.
Focus: $5k-6=0 \implies k=6/5$. $5h+2 = 3a \implies h = \frac{3a-2}{5}$.
Given the options,there might be a typo in the question's point $A$. If $A=(-2, 3)$,then $h = \frac{3at^2-4}{5}$,$k = \frac{6at+6}{5}$.
$5h+4 = 3at^2$,$5k-6 = 6at \implies (5k-6)^2 = 36a^2t^2 = 36a \left(\frac{5h+4}{3}\right) = 12a(5h+4)$.
Focus: $5k-6=0 \implies k=6/5$. $5h+4 = 3a \implies h = \frac{3a-4}{5}$.
This matches option $C$.

(B) Let the four points be $A(1, 1, 1)$,$B(-2, 3, 2)$,$C(x, -5, 3)$,and $D(1, y, -1)$.
Since the points are coplanar,the vectors $\vec{AB}$,$\vec{AC}$,and $\vec{AD}$ must be coplanar,which implies their scalar triple product is zero:
$\vec{AB} = (-2-1)\hat{i} + (3-1)\hat{j} + (2-1)\hat{k} = -3\hat{i} + 2\hat{j} + \hat{k}$
$\vec{AC} = (x-1)\hat{i} + (-5-1)\hat{j} + (3-1)\hat{k} = (x-1)\hat{i} - 6\hat{j} + 2\hat{k}$
$\vec{AD} = (1-1)\hat{i} + (y-1)\hat{j} + (-1-1)\hat{k} = 0\hat{i} + (y-1)\hat{j} - 2\hat{k}$
Setting the determinant to zero:
$\begin{vmatrix} -3 & 2 & 1 \\ x-1 & -6 & 2 \\ 0 & y-1 & -2 \end{vmatrix} = 0$
Expanding along the first column:
$-3[(-6)(-2) - 2(y-1)] - (x-1)[(2)(-2) - (1)(y-1)] + 0 = 0$
$-3[12 - 2y + 2] - (x-1)[-4 - y + 1] = 0$
$-3[14 - 2y] - (x-1)[-y - 3] = 0$
$-42 + 6y + (x-1)(y+3) = 0$
$-42 + 6y + xy + 3x - y - 3 = 0$
$xy + 3x + 5y - 45 = 0$
Adding $15$ to both sides to factorize:
$xy + 3x + 5y + 15 = 45 + 15$
$x(y+3) + 5(y+3) = 60$
$(x+5)(y+3) = 60$

(D) Given points are $A(1, -2, -1)$,$B(4, 0, -3)$,$C(1, 2, -1)$,and $D(2, -4, -5)$.
The vectors are:
$\vec{b} = \vec{AB} = (4-1)\hat{i} + (0-(-2))\hat{j} + (-3-(-1))\hat{k} = 3\hat{i} + 2\hat{j} - 2\hat{k}$
$\vec{c} = \vec{AC} = (1-1)\hat{i} + (2-(-2))\hat{j} + (-1-(-1))\hat{k} = 0\hat{i} + 4\hat{j} + 0\hat{k}$
$\vec{d} = \vec{AD} = (2-1)\hat{i} + (-4-(-2))\hat{j} + (-5-(-1))\hat{k} = 1\hat{i} - 2\hat{j} - 4\hat{k}$
The scalar triple product $[\vec{b}, \vec{c}, \vec{d}] = \begin{vmatrix} 3 & 2 & -2 \\ 0 & 4 & 0 \\ 1 & -2 & -4 \end{vmatrix} = 3(-16) - 2(0) - 2(-4) = -48 + 8 = -40$.
We know that $[\vec{b} \times \vec{c}, \vec{c} \times \vec{d}, \vec{d} \times \vec{b}] = [\vec{b}, \vec{c}, \vec{d}]^2 = (-40)^2 = 1600$.
Also,$[\vec{b}+\vec{c}, \vec{c}+\vec{d}, \vec{d}+\vec{b}] = 2[\vec{b}, \vec{c}, \vec{d}] = 2(-40) = -80$.
Therefore,the required value is $\frac{1600}{-80} = -20$.

(D) Let the position vectors be $\vec{a} = P \hat{i} - 2 \hat{j} + 3 \hat{k}$,$\vec{b} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$,and $\vec{c} = 4 \hat{i} + 13 \hat{j} - 18 \hat{k}$.
Since the points $A$,$B$,and $C$ are collinear,the vectors $\vec{AB}$ and $\vec{BC}$ must be parallel.
$\vec{AB} = \vec{b} - \vec{a} = (2-P) \hat{i} + 5 \hat{j} - 7 \hat{k}$
$\vec{BC} = \vec{c} - \vec{b} = (4-2) \hat{i} + (13-3) \hat{j} + (-18+4) \hat{k} = 2 \hat{i} + 10 \hat{j} - 14 \hat{k}$
Since $\vec{AB} = k \vec{BC}$,we have $(2-P) \hat{i} + 5 \hat{j} - 7 \hat{k} = k(2 \hat{i} + 10 \hat{j} - 14 \hat{k})$.
Comparing the coefficients of $\hat{j}$,we get $5 = 10k \Rightarrow k = 0.5$.
Comparing the coefficients of $\hat{i}$,we get $2-P = 2k = 2(0.5) = 1 \Rightarrow P = 1$.
Thus,$\vec{AB} = (2-1) \hat{i} + 5 \hat{j} - 7 \hat{k} = \hat{i} + 5 \hat{j} - 7 \hat{k}$.
The magnitude $|\vec{AB}| = \sqrt{1^2 + 5^2 + (-7)^2} = \sqrt{1 + 25 + 49} = \sqrt{75} = 5 \sqrt{3}$.
The unit vector in the direction of $\vec{AB}$ is $\hat{u} = \frac{\vec{AB}}{|\vec{AB}|} = \frac{1}{5 \sqrt{3}} (\hat{i} + 5 \hat{j} - 7 \hat{k})$.
Since $|P| = |1| = 1$,the required vector is $1 \times \hat{u} = \frac{1}{5 \sqrt{3}} (\hat{i} + 5 \hat{j} - 7 \hat{k})$.

(B) Given,$\vec{b}=2 \hat{i}-\hat{j}-\hat{k}$ and $\vec{a}=3 \hat{i}+4 \hat{j}-5 \hat{k}$.
First,calculate the dot products:
$\vec{b} \cdot \vec{b} = (2)^2 + (-1)^2 + (-1)^2 = 4 + 1 + 1 = 6$.
$\vec{b} \cdot \vec{a} = (2)(3) + (-1)(4) + (-1)(-5) = 6 - 4 + 5 = 7$.
Using the vector triple product formula $\vec{b} \times(\vec{a} \times \vec{b})=(\vec{b} \cdot \vec{b}) \vec{a}-(\vec{b} \cdot \vec{a}) \vec{b}$,we get:
$\vec{b} \times(\vec{a} \times \vec{b}) = 6 \vec{a} - 7 \vec{b}$.
We can rewrite this as $\frac{\vec{a} - (7/6) \vec{b}}{1/6}$.
Comparing this with the given expression $\frac{\vec{a}-k \vec{b}}{l}$,we find $k = 7/6$ and $l = 1/6$.
Now,calculate $|\vec{b}| = \sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{6}$.
Finally,$\frac{k}{l|\vec{b}|} = \frac{7/6}{(1/6) \times \sqrt{6}} = \frac{7}{\sqrt{6}}$.
This value represents the scalar projection of $\vec{a}$ on $\vec{b}$,which is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{7}{\sqrt{6}}$.

(A) Given that $a, b, c$ are unit vectors,so $|a| = |b| = |c| = 1$.
Using the vector triple product formula: $a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$.
Given $a \times (b \times c) = \frac{1}{2}b$,we have:
$(a \cdot c)b - (a \cdot b)c = \frac{1}{2}b$.
Since $b$ and $c$ are generally non-collinear,we compare the coefficients:
$a \cdot c = \frac{1}{2}$ and $a \cdot b = 0$.
For $a \cdot c = \frac{1}{2}$,we have $|a||c| \cos \theta_2 = \frac{1}{2} \Rightarrow (1)(1) \cos \theta_2 = \frac{1}{2} \Rightarrow \theta_2 = 60^{\circ}$.
For $a \cdot b = 0$,we have $|a||b| \cos \theta_1 = 0 \Rightarrow (1)(1) \cos \theta_1 = 0 \Rightarrow \theta_1 = 90^{\circ}$.
Therefore,$\theta_1 + \theta_2 = 90^{\circ} + 60^{\circ} = 150^{\circ}$.

(D) The centroid of a tetrahedron with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3)$ and $(x_4, y_4, z_4)$ is given by $\left(\frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4}, \frac{z_1+z_2+z_3+z_4}{4}\right)$.
Given vertices are $(a, 2, 1), (1, b, 4), (4, 0, c)$ and $(1, 1, 7)$.
So,the centroid is $\left(\frac{a+1+4+1}{4}, \frac{2+b+0+1}{4}, \frac{1+4+c+7}{4}\right) = \left(\frac{a+6}{4}, \frac{b+3}{4}, \frac{c+12}{4}\right)$.
Equating this to the given centroid $\left(\frac{9}{4}, \frac{5}{4}, \frac{15}{4}\right)$:
For $x$-coordinate: $\frac{a+6}{4} = \frac{9}{4} \Rightarrow a+6 = 9 \Rightarrow a = 3$.
For $y$-coordinate: $\frac{b+3}{4} = \frac{5}{4} \Rightarrow b+3 = 5 \Rightarrow b = 2$.
For $z$-coordinate: $\frac{c+12}{4} = \frac{15}{4} \Rightarrow c+12 = 15 \Rightarrow c = 3$.
Thus,$a=3, b=2, c=3$. Comparing these values,we get $a=c=b+1$ (since $3=3=2+1$).

(D) Let the point $A$ with position vector $\vec{a} = \frac{-9}{2} \bar{i} - 6 \bar{j} + \frac{1}{2} \bar{k}$ divide the line segment $PQ$ in the ratio $\lambda : 1$.
Using the section formula,the position vector of $A$ is given by:
$\vec{a} = \frac{\lambda \vec{q} + 1 \vec{p}}{\lambda + 1}$
Substituting the given vectors $\vec{p} = -2 \bar{i} - 3 \bar{j} + \bar{k}$ and $\vec{q} = 3 \bar{i} + 3 \bar{j} + 2 \bar{k}$:
$\frac{-9}{2} \bar{i} - 6 \bar{j} + \frac{1}{2} \bar{k} = \frac{\lambda(3 \bar{i} + 3 \bar{j} + 2 \bar{k}) + 1(-2 \bar{i} - 3 \bar{j} + \bar{k})}{\lambda + 1}$
Comparing the $x$-coordinates:
$\frac{-9}{2} = \frac{3 \lambda - 2}{\lambda + 1}$
$-9(\lambda + 1) = 2(3 \lambda - 2)$
$-9 \lambda - 9 = 6 \lambda - 4$
$-15 \lambda = 5$
$\lambda = -\frac{5}{15} = -\frac{1}{3}$
Since $\lambda$ is negative,the point $A$ divides the line segment $PQ$ externally in the ratio $1 : 3$.

(A) Since points $A(1, x, 3)$,$B(3, 4, 7)$,and $C(y, -2, -5)$ are collinear,the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ must be parallel.\\
$\overrightarrow{AB} = (3-1)\hat{i} + (4-x)\hat{j} + (7-3)\hat{k} = 2\hat{i} + (4-x)\hat{j} + 4\hat{k}$\\
$\overrightarrow{BC} = (y-3)\hat{i} + (-2-4)\hat{j} + (-5-7)\hat{k} = (y-3)\hat{i} - 6\hat{j} - 12\hat{k}$\\
Since $\overrightarrow{AB} = k \overrightarrow{BC}$ for some scalar $k$,we have:\\
$2 = k(y-3)$\\
$4-x = k(-6)$\\
$4 = k(-12)$\\
From the third equation,$k = \frac{4}{-12} = -\frac{1}{3}$.\\
Substituting $k = -\frac{1}{3}$ into the first equation: $2 = -\frac{1}{3}(y-3) \Rightarrow -6 = y-3 \Rightarrow y = -3$.\\
Substituting $k = -\frac{1}{3}$ into the second equation: $4-x = -\frac{1}{3}(-6) \Rightarrow 4-x = 2 \Rightarrow x = 2$.\\
Therefore,$x+y = 2 + (-3) = -1$.

(C) Let the points be $A(2, 3, -1)$ and $O(0, 0, 0)$.
The direction ratios of the line $OA$ are $(2-0, 3-0, -1-0) = (2, 3, -1)$.
The distance $OA$ is $\sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$.
The direction cosines $(l, m, n)$ are given by $\frac{a}{r}, \frac{b}{r}, \frac{c}{r}$,where $(a, b, c)$ are direction ratios and $r$ is the distance.
Thus,$l = \frac{2}{\sqrt{14}}, m = \frac{3}{\sqrt{14}}, n = \frac{-1}{\sqrt{14}}$.
Alternatively,for the line $AO$,the direction ratios are $(0-2, 0-3, 0-(-1)) = (-2, -3, 1)$.
The direction cosines are $\frac{-2}{\sqrt{14}}, \frac{-3}{\sqrt{14}}, \frac{1}{\sqrt{14}}$.
Comparing with the options,option $C$ is correct.

(B) Let the direction cosines of the line perpendicular to both $L_1$ and $L_2$ be $(l, m, n)$.
Since the line is perpendicular to $L_1$ with direction ratios $(1, -2, -2)$,we have:
$l - 2m - 2n = 0$ $(i)$
Since the line is perpendicular to $L_2$ with direction ratios $(0, 2, 1)$,we have:
$0l + 2m + n = 0 \Rightarrow n = -2m$ $(ii)$
Substituting $(ii)$ into $(i)$:
$l - 2m - 2(-2m) = 0$
$l - 2m + 4m = 0$
$l + 2m = 0 \Rightarrow l = -2m$
We know that for direction cosines,$l^2 + m^2 + n^2 = 1$.
Substituting $l = -2m$ and $n = -2m$:
$(-2m)^2 + m^2 + (-2m)^2 = 1$
$4m^2 + m^2 + 4m^2 = 1$
$9m^2 = 1 \Rightarrow m^2 = \frac{1}{9} \Rightarrow |m| = \frac{1}{3}$
Since $l = -2m$,$|l| = |-2m| = 2|m| = 2(\frac{1}{3}) = \frac{2}{3}$.
Since $n = -2m$,$|n| = |-2m| = 2|m| = 2(\frac{1}{3}) = \frac{2}{3}$.
Therefore,$|l| + |m| + |n| = \frac{2}{3} + \frac{1}{3} + \frac{2}{3} = \frac{5}{3}$.

(C) Given the direction ratios of two lines are $(a_1, b_1, c_1) = (2, -1, 2)$ and $(a_2, b_2, c_2) = (K, -3, -5)$,and the angle between them is $\theta = 60^{\circ}$.
Using the formula for the angle between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$:
$\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$
Substituting the given values:
$\cos 60^{\circ} = \frac{|2K + (-1)(-3) + 2(-5)|}{\sqrt{2^2 + (-1)^2 + 2^2} \sqrt{K^2 + (-3)^2 + (-5)^2}}$
$\frac{1}{2} = \frac{|2K + 3 - 10|}{\sqrt{4 + 1 + 4} \sqrt{K^2 + 9 + 25}}$
$\frac{1}{2} = \frac{|2K - 7|}{3 \sqrt{K^2 + 34}}$
$3 \sqrt{K^2 + 34} = 2 |2K - 7|$
Squaring both sides:
$9(K^2 + 34) = 4(2K - 7)^2$
$9K^2 + 306 = 4(4K^2 - 28K + 49)$
$9K^2 + 306 = 16K^2 - 112K + 196$
$7K^2 - 112K - 110 = 0$

(C) The line $OP$ passes through the origin $O(0, 0, 0)$ and has direction ratios $3, -2, 6$.
Thus,the coordinates of any point $P$ on this line can be written as $(3\lambda, -2\lambda, 6\lambda)$ for some scalar $\lambda$.
The distance of point $P$ from the origin $O$ is given by $|OP| = \sqrt{(3\lambda)^2 + (-2\lambda)^2 + (6\lambda)^2}$.
Given that $|OP| = 63$,we have:
$\sqrt{9\lambda^2 + 4\lambda^2 + 36\lambda^2} = 63$
$\sqrt{49\lambda^2} = 63$
$7|\lambda| = 63$
$|\lambda| = 9$
Taking $\lambda = 9$,the coordinates of $P$ are $(3(9), -2(9), 6(9)) = (27, -18, 54)$.
Taking $\lambda = -9$,the coordinates of $P$ are $(-27, 18, -54)$.
Comparing with the given options,the correct coordinates are $(27, -18, 54)$.

(A) Let the vector be $\vec{a} = \hat{i} + \hat{j} - 2\hat{k}$.
Its direction ratios are $(a, b, c) = (1, 1, -2)$.
The magnitude of the vector is $|\vec{a}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.
The direction cosines $(l, m, n)$ are given by $\left(\frac{a}{|\vec{a}|}, \frac{b}{|\vec{a}|}, \frac{c}{|\vec{a}|}\right)$.
Substituting the values,we get $\left(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{-2}{\sqrt{6}}\right)$.

(C) The direction ratios of the normal to the plane $AOB$ are given by the cross product of the vectors representing the rays $OA$ and $OB$. Let $\vec{a} = \langle 1, 2, 3 \rangle$ and $\vec{b} = \langle -1, 0, 1 \rangle$. The normal vector $\vec{n}$ is given by $\vec{a} \times \vec{b}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -1 & 0 & 1 \end{vmatrix} = \hat{i}(2-0) - \hat{j}(1+3) + \hat{k}(0+2) = 2\hat{i} - 4\hat{j} + 2\hat{k}$.
Thus,the direction ratios of the normal are $\langle 2, -4, 2 \rangle$,which can be simplified to $\langle 1, -2, 1 \rangle$ or $\langle -1, 2, -1 \rangle$.
The magnitude of the vector $\langle -1, 2, -1 \rangle$ is $\sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$.
The direction cosines are $\frac{-1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}$.

(C) The shortest distance $d$ between two skew lines $\vec{r}=\vec{a}_1+\lambda \vec{b}_1$ and $\vec{r}=\vec{a}_2+\mu \vec{b}_2$ is given by the formula: $d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|$.
Given lines are $\vec{r}=(3 \hat{i}+4 \hat{j}-2 \hat{k})+\lambda(-\hat{i}+2 \hat{j}+\hat{k})$ and $\vec{r}=(\hat{i}-7 \hat{j}-2 \hat{k})+\mu(\hat{i}+3 \hat{j}+2 \hat{k})$.
Here,$\vec{a}_1 = 3 \hat{i}+4 \hat{j}-2 \hat{k}$,$\vec{b}_1 = -\hat{i}+2 \hat{j}+\hat{k}$,$\vec{a}_2 = \hat{i}-7 \hat{j}-2 \hat{k}$,and $\vec{b}_2 = \hat{i}+3 \hat{j}+2 \hat{k}$.
First,calculate $\vec{a}_2 - \vec{a}_1 = (1-3)\hat{i} + (-7-4)\hat{j} + (-2-(-2))\hat{k} = -2\hat{i} - 11\hat{j}$.
Next,calculate the cross product $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 1 \\ 1 & 3 & 2 \end{vmatrix} = \hat{i}(4-3) - \hat{j}(-2-1) + \hat{k}(-3-2) = \hat{i} + 3\hat{j} - 5\hat{k}$.
The magnitude $|\vec{b}_1 \times \vec{b}_2| = \sqrt{1^2 + 3^2 + (-5)^2} = \sqrt{1+9+25} = \sqrt{35}$.
Now,the dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-2\hat{i} - 11\hat{j} + 0\hat{k}) \cdot (\hat{i} + 3\hat{j} - 5\hat{k}) = (-2)(1) + (-11)(3) + (0)(-5) = -2 - 33 = -35$.
Finally,$d = \left| \frac{-35}{\sqrt{35}} \right| = \sqrt{35}$.

(C) Two lines $\vec{r}=\vec{a}_1+t\vec{b}_1$ and $\vec{r}=\vec{a}_2+s\vec{b}_2$ are parallel if $\vec{b}_1=m\vec{b}_2$ for some scalar $m \in \mathbb{R}$.
They are perpendicular if $\vec{b}_1 \cdot \vec{b}_2 = 0$.
They intersect if the shortest distance between them is $0$.
Given $L_1: \vec{r}=(\hat{i}+5 \hat{j}+5 \hat{k})+t(4 \hat{i}-4 \hat{j}+5 \hat{k})$ and $L_2: \vec{r}=(2 \hat{i}+4 \hat{j}+5 \hat{k})+s(8 \hat{i}-3 \hat{j}+\hat{k})$.
Here,$\vec{b}_1 = 4\hat{i}-4\hat{j}+5\hat{k}$ and $\vec{b}_2 = 8\hat{i}-3\hat{j}+\hat{k}$.
Since $\vec{b}_1$ is not a scalar multiple of $\vec{b}_2$,the lines are not parallel.
Check perpendicularity: $\vec{b}_1 \cdot \vec{b}_2 = (4)(8) + (-4)(-3) + (5)(1) = 32 + 12 + 5 = 49 \neq 0$. Thus,they are not perpendicular.
To check if they are skew,we calculate the shortest distance $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$.
$\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (4-5)\hat{j} + (5-5)\hat{k} = \hat{i} - \hat{j}$.
$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -4 & 5 \\ 8 & -3 & 1 \end{vmatrix} = \hat{i}(-4+15) - \hat{j}(4-40) + \hat{k}(-12+32) = 11\hat{i} + 36\hat{j} + 20\hat{k}$.
$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (1)(11) + (-1)(36) + (0)(20) = 11 - 36 = -25 \neq 0$.
Since the shortest distance is non-zero,the lines are skew lines.

(A) The equation of the line passing through the point $A(2, 1, 0)$ and parallel to the vector $\vec{v} = \hat{i} + 5\hat{j} + 2\hat{k}$ is given by $\frac{x-2}{1} = \frac{y-1}{5} = \frac{z-0}{2} = t$.
Any point $R$ on this line can be represented as $R(t+2, 5t+1, 2t)$.
Let $P(3, 5, 2)$ be the given point. The vector $\vec{PR}$ is given by $\vec{PR} = (t+2-3)\hat{i} + (5t+1-5)\hat{j} + (2t-2)\hat{k} = (t-1)\hat{i} + (5t-4)\hat{j} + (2t-2)\hat{k}$.
Since $\vec{PR}$ is perpendicular to the line,its dot product with the direction vector $\vec{v} = \hat{i} + 5\hat{j} + 2\hat{k}$ must be zero:
$(t-1)(1) + (5t-4)(5) + (2t-2)(2) = 0$.
$t - 1 + 25t - 20 + 4t - 4 = 0$.
$30t - 25 = 0 \implies t = \frac{25}{30} = \frac{5}{6}$.
Substituting $t = \frac{5}{6}$ into the coordinates of $R$,we get $R = (\frac{5}{6}+2, 5(\frac{5}{6})+1, 2(\frac{5}{6})) = (\frac{17}{6}, \frac{31}{6}, \frac{10}{6})$.
The perpendicular distance $d$ is the magnitude of vector $\vec{PR}$:
$d = \sqrt{(\frac{17}{6}-3)^2 + (\frac{31}{6}-5)^2 + (\frac{10}{6}-2)^2} = \sqrt{(-\frac{1}{6})^2 + (\frac{1}{6})^2 + (-\frac{2}{6})^2} = \sqrt{\frac{1}{36} + \frac{1}{36} + \frac{4}{36}} = \sqrt{\frac{6}{36}} = \frac{\sqrt{6}}{6} = \frac{1}{\sqrt{6}}$.

(D) Given the two skew lines are $r = b + ta$ and $r = d + sc$.
The formula for the shortest distance between two skew lines $r = a_1 + \lambda b_1$ and $r = a_2 + \mu b_2$ is given by $d = \left| \frac{(a_2 - a_1) \cdot (b_1 \times b_2)}{|b_1 \times b_2|} \right|$.
Here,$a_1 = b$,$b_1 = a$,$a_2 = d$,and $b_2 = c$.
Substituting these values,the shortest distance is $\left| \frac{(d - b) \cdot (a \times c)}{|a \times c|} \right|$.
Since $|x| = |-x|$,this is equivalent to $\left| \frac{(b - d) \cdot (a \times c)}{|a \times c|} \right|$.
This expression represents the magnitude of the orthogonal projection of the vector $(b - d)$ onto the vector $(a \times c)$.

(D) The normal vectors to the planes $2x-2y+z=0$ and $x-y+2z=4$ are $\vec{n_1} = (2, -2, 1)$ and $\vec{n_2} = (1, -1, 2)$ respectively.
The normal vector to the plane $ax+by+cz+1=0$ is $\vec{n} = (a, b, c)$.
Since the plane is perpendicular to the given two planes,its normal vector $\vec{n}$ must be parallel to the cross product of $\vec{n_1}$ and $\vec{n_2}$.
$\vec{n} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & -1 & 2 \end{vmatrix} = \hat{i}(-4+1) - \hat{j}(4-1) + \hat{k}(-2+2) = -3\hat{i} - 3\hat{j} + 0\hat{k}$.
Thus,the normal vector is proportional to $(-3, -3, 0)$,which simplifies to $(1, 1, 0)$.
The equation of the plane is $1(x-1) + 1(y+2) + 0(z-1) = 0$,which simplifies to $x+y+1=0$.
Comparing this with $ax+by+cz+1=0$,we get $a=1, b=1, c=0$.
Therefore,$a+b-c = 1+1-0 = 2$.

(A) Given,distance from the origin to the plane $d = 6$ units.
Normal vector $\vec{N} = 2 \hat{i} + 6 \hat{j} - 3 \hat{k}$.
First,find the magnitude of the normal vector: $|\vec{N}| = \sqrt{2^2 + 6^2 + (-3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7$.
The unit normal vector is $\hat{n} = \frac{\vec{N}}{|\vec{N}|} = \frac{2 \hat{i} + 6 \hat{j} - 3 \hat{k}}{7}$.
The equation of the plane in normal form is $\vec{r} \cdot \hat{n} = d$.
Substituting the values,we get $\vec{r} \cdot \left( \frac{2 \hat{i} + 6 \hat{j} - 3 \hat{k}}{7} \right) = 6$.
Multiplying by $7$,we get $\vec{r} \cdot (2 \hat{i} + 6 \hat{j} - 3 \hat{k}) = 42$.
Substituting $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$,we get $(x \hat{i} + y \hat{j} + z \hat{k}) \cdot (2 \hat{i} + 6 \hat{j} - 3 \hat{k}) = 42$.
This simplifies to $2x + 6y - 3z = 42$,or $2x + 6y - 3z - 42 = 0$.

(C) Given vectors are $\vec{a}=2 \hat{i}-2 \hat{j}+5 \hat{k}$ and $\vec{b}=-2 \hat{i}+5 \hat{j}-3 \hat{k}$.
Sum of the vectors is $\vec{s} = \vec{a} + \vec{b} = (2-2) \hat{i} + (-2+5) \hat{j} + (5-3) \hat{k} = 0 \hat{i} + 3 \hat{j} + 2 \hat{k} = 3 \hat{j} + 2 \hat{k}$.
Since the $\hat{i}$ component of the resulting vector $\vec{s}$ is $0$,the vector lies in the $YZ$-plane.
$A$ vector that lies in a plane is parallel to that plane.
Therefore,the vector is parallel to the $YZ$-plane.

(B) The given plane is $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=1$.
Converting this to Cartesian form,we get $2 x+3 y-4 z=1$,or $2 x+3 y-4 z-1=0$.
Any plane parallel to this plane is of the form $2 x+3 y-4 z+\lambda=0$.
The distance $d$ between two parallel planes $Ax+By+Cz+D_1=0$ and $Ax+By+Cz+D_2=0$ is given by $d = \frac{|D_1-D_2|}{\sqrt{A^2+B^2+C^2}}$.
Here,$d=2$,$A=2$,$B=3$,$C=-4$,$D_1=-1$,and $D_2=\lambda$.
So,$2 = \frac{|\lambda-(-1)|}{\sqrt{2^2+3^2+(-4)^2}} = \frac{|\lambda+1|}{\sqrt{4+9+16}} = \frac{|\lambda+1|}{\sqrt{29}}$.
This implies $|\lambda+1| = 2 \sqrt{29}$,so $\lambda+1 = \pm 2 \sqrt{29}$,which means $\lambda = -1 \pm 2 \sqrt{29}$.
Substituting $\lambda$ back into the equation $2 x+3 y-4 z+\lambda=0$,we get $2 x+3 y-4 z-1 \pm 2 \sqrt{29} = 0$,or $2 x+3 y-4 z = 1 \mp 2 \sqrt{29}$.
Since the options provide $1 \pm 2 \sqrt{29}$,the correct equation is $2 x+3 y-4 z = 1 \pm 2 \sqrt{29}$.

(B) The equation of the plane passing through points $A(1, 2, 3)$,$B(2, 3, -4)$,and $C(3, -4, 5)$ is given by the determinant equation:
$\left|\begin{array}{ccc} x-1 & y-2 & z-3 \\ 2-1 & 3-2 & -4-3 \\ 3-1 & -4-2 & 5-3 \end{array}\right| = 0$
$\Rightarrow \left|\begin{array}{ccc} x-1 & y-2 & z-3 \\ 1 & 1 & -7 \\ 2 & -6 & 2 \end{array}\right| = 0$
Expanding along the first row:
$(x-1)(2 - 42) - (y-2)(2 - (-14)) + (z-3)(-6 - 2) = 0$
$(x-1)(-40) - (y-2)(16) + (z-3)(-8) = 0$
$-40x + 40 - 16y + 32 - 8z + 24 = 0$
$-40x - 16y - 8z + 96 = 0$
Dividing by $-8$:
$5x + 2y + z - 12 = 0$
The perpendicular distance $d$ from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.
$d = \frac{|-12|}{\sqrt{5^2 + 2^2 + 1^2}} = \frac{12}{\sqrt{25 + 4 + 1}} = \frac{12}{\sqrt{30}}$.

(A) The equation of a plane passing through three points $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$ is given by the determinant equation:
$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0$
Substituting the given points $A(1,1,-1)$,$B(2,-1,0)$,and $C(-1,0,2)$:
$\begin{vmatrix} x-1 & y-1 & z+1 \\ 2-1 & -1-1 & 0+1 \\ -1-1 & 0-1 & 2+1 \end{vmatrix} = 0$
$\begin{vmatrix} x-1 & y-1 & z+1 \\ 1 & -2 & 1 \\ -2 & -1 & 3 \end{vmatrix} = 0$
Expanding along the first row:
$(x-1)(-6+1) - (y-1)(3+2) + (z+1)(-1-4) = 0$
$-5(x-1) - 5(y-1) - 5(z+1) = 0$
Dividing by $-5$:
$(x-1) + (y-1) + (z+1) = 0$
$x + y + z - 1 = 0$
Now,check the options by substituting the coordinates into the equation $x + y + z - 1 = 0$:
For $(1,2,-2)$: $1 + 2 - 2 - 1 = 0$. This satisfies the equation.
Thus,the point $(1,2,-2)$ lies on the plane.

(B) The normal vector $\vec{n}$ to the plane is given by the cross product of the two parallel vectors $\vec{v_1} = 2\hat{i}+3\hat{j}+\hat{k}$ and $\vec{v_2} = -\hat{i}+2\hat{j}-3\hat{k}$.
$\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ -1 & 2 & -3 \end{vmatrix} = \hat{i}(-9-2) - \hat{j}(-6+1) + \hat{k}(4+3) = -11\hat{i} + 5\hat{j} + 7\hat{k}$.
The equation of the plane passing through $(1, 2, 1)$ with normal vector $\vec{n} = -11\hat{i} + 5\hat{j} + 7\hat{k}$ is:
$-11(x-1) + 5(y-2) + 7(z-1) = 0$
$-11x + 11 + 5y - 10 + 7z - 7 = 0$
$-11x + 5y + 7z - 6 = 0$
$-11x + 5y + 7z = 6$
Dividing by $6$,we get:
$-\frac{11}{6}x + \frac{5}{6}y + \frac{7}{6}z = 1$
Comparing this with $ax+by+cz=1$,we have $a = -\frac{11}{6}$,$b = \frac{5}{6}$,$c = \frac{7}{6}$.
Therefore,$18(a+b+c) = 18 \left(-\frac{11}{6} + \frac{5}{6} + \frac{7}{6}\right) = 18 \left(\frac{1}{6}\right) = 3$.

(A) The equation of a plane passing through $(2,-3,4)$ with normal vector $\vec{n} = (a, b, c)$ is given by $a(x-2) + b(y+3) + c(z-4) = 0$,which simplifies to $ax + by + cz - 2a + 3b - 4c = 0$ ... $(i)$
Since the plane is perpendicular to both $2x-3y+5z=2$ and $x+y+2z=3$,its normal vector $\vec{n}$ must be parallel to the cross product of the normals of the given planes,$\vec{n_1} = (2, -3, 5)$ and $\vec{n_2} = (1, 1, 2)$.
$\vec{n} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 5 \\ 1 & 1 & 2 \end{vmatrix} = \hat{i}(-6-5) - \hat{j}(4-5) + \hat{k}(2+3) = -11\hat{i} + 1\hat{j} + 5\hat{k}$.
Thus,$(a, b, c) = (-11, 1, 5)$.
Substituting these into Eq. $(i)$: $-11x + y + 5z - 2(-11) + 3(1) - 4(5) = 0$
$-11x + y + 5z + 22 + 3 - 20 = 0$
$-11x + y + 5z + 5 = 0 \Rightarrow 11x - y - 5z = 5$
Dividing by $11$,we get $x - \frac{1}{11}y - \frac{5}{11}z = \frac{5}{11}$.
Comparing this with $x + py + qz = r$,we find $r = \frac{5}{11}$.

(D) Given the vector equation of the plane: $\vec{r} = (2+\alpha-3\beta) \hat{i} + (\beta-3) \hat{j} + (2\alpha-5\beta-1) \hat{k}$.
Let $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$.
Comparing the components,we have:
$x = 2 + \alpha - 3\beta$ $(i)$
$y = \beta - 3 \implies \beta = y + 3$ (ii)
$z = 2\alpha - 5\beta - 1$ (iii)
Substitute $\beta = y + 3$ into (iii):
$z = 2\alpha - 5(y + 3) - 1$
$z = 2\alpha - 5y - 15 - 1$
$z = 2\alpha - 5y - 16$
$2\alpha = z + 5y + 16 \implies \alpha = \frac{z + 5y + 16}{2}$.
Now,substitute $\alpha$ and $\beta$ into $(i)$:
$x = 2 + \left(\frac{z + 5y + 16}{2}\right) - 3(y + 3)$
Multiply by $2$:
$2x = 4 + z + 5y + 16 - 6y - 18$
$2x = z - y + 2$
$2x + y - z = 2$.

(B) Given the matrix equation: $\begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 5 \\ -5 \\ 10 \end{bmatrix}$
Multiplying the first two matrices: $\begin{bmatrix} 1(1) & 1(-1) & 1(2) \\ -1(1) & -1(-1) & -1(2) \\ 2(1) & 2(-1) & 2(2) \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 5 \\ -5 \\ 10 \end{bmatrix}$
$\Rightarrow \begin{bmatrix} 1 & -1 & 2 \\ -1 & 1 & -2 \\ 2 & -2 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 5 \\ -5 \\ 10 \end{bmatrix}$
This results in the system of linear equations:
$x - y + 2z = 5$
$-x + y - 2z = -5$
$2x - 2y + 4z = 10$
All three equations are equivalent to the single plane equation $x - y + 2z = 5$.
Since the coefficients of $x, y, z$ are all non-zero,the plane is not parallel to any of the coordinate planes $(XY, YZ, ZX)$.
Thus,the solutions lie on a plane not parallel to any of the coordinate planes.

(A) Given the line $\vec{r}=\vec{a}+2 \vec{b}+p(\vec{a}-2 \vec{c}) \quad \dots(1)$ and the plane $\vec{r}=3 \vec{a}-q(\vec{c}-\vec{b})+k(\vec{a}-\vec{b}+\vec{c}) \quad \dots(2)$.
Equating the two expressions for $\vec{r}$:
$\vec{a}+2 \vec{b}+p \vec{a}-2p \vec{c} = 3 \vec{a}-q \vec{c}+q \vec{b}+k \vec{a}-k \vec{b}+k \vec{c}$
$\vec{a}(1+p) + 2 \vec{b} - 2p \vec{c} = \vec{a}(3+k) + \vec{b}(q-k) + \vec{c}(k-q)$
Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar,we compare the coefficients:
$1+p = 3+k \Rightarrow p-k = 2$
$2 = q-k \Rightarrow q-k = 2$
$-2p = k-q \Rightarrow q-k = 2p$
From $q-k=2$ and $q-k=2p$,we get $2p=2 \Rightarrow p=1$.
Substituting $p=1$ into $p-k=2$,we get $1-k=2 \Rightarrow k=-1$.
Substituting $k=-1$ into $q-k=2$,we get $q-(-1)=2 \Rightarrow q=1$.
Now,substitute $p=1$ into equation $(1)$ to find the position vector $\vec{r}$:
$\vec{r} = \vec{a}+2 \vec{b}+1(\vec{a}-2 \vec{c}) = 2 \vec{a}+2 \vec{b}-2 \vec{c}$.
Comparing this with $\vec{r}=x \vec{a}+y \vec{b}+z \vec{c}$,we get $x=2, y=2, z=-2$.
Therefore,$x y z = 2 \times 2 \times (-2) = -8$.

(B) Given $P(A / B) = \frac{P(A \cap B)}{P(B)} = \frac{3}{10} \implies P(A \cap B) = \frac{3}{10} P(B)$.
$P(B / A) = \frac{P(A \cap B)}{P(A)} = \frac{4}{5} \implies P(A) = \frac{5}{4} P(A \cap B) = \frac{5}{4} \times \frac{3}{10} P(B) = \frac{3}{8} P(B)$.
We know $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Given $P(A \cup B) = K P(B)$,so $K P(B) = P(A) + P(B) - P(A \cap B)$.
Dividing by $P(B)$,we get $K = \frac{P(A)}{P(B)} + 1 - \frac{P(A \cap B)}{P(B)}$.
Substituting the values: $K = \frac{3}{8} + 1 - \frac{3}{10}$.
$K = \frac{15 + 40 - 12}{40} = \frac{43}{40}$.
Therefore,$\frac{1}{K} = \frac{40}{43}$.

(A) bag $B$ contains $4$ white balls and $2$ black balls. Total balls in bag $B = 4 + 2 = 6$.
Probability of selecting $1$ white ball from bag $B$ is $P_1 = \frac{4}{6} = \frac{2}{3}$.
Another bag $C$ contains $3$ white balls and $5$ black balls. Total balls in bag $C = 3 + 5 = 8$.
Probability of selecting $1$ white ball from bag $C$ is $P_2 = \frac{3}{8}$.
Since the events are independent,the probability that both balls drawn are white is $P = P_1 \times P_2 = \frac{2}{3} \times \frac{3}{8} = \frac{6}{24} = \frac{1}{4}$.

(C) Let $E$ be the event of selecting a science book.
Let $A$ be the event of selecting a book from group $A$,and $B$ be the event of selecting a book from group $B$.
The probability of getting $2$ or $5$ on a die is $P(A) = \frac{2}{6} = \frac{1}{3}$.
The probability of not getting $2$ or $5$ is $P(B) = 1 - \frac{1}{3} = \frac{2}{3}$.
In group $A$,there are $8$ science and $5$ engineering books,total $13$ books. So,$P(E|A) = \frac{8}{13}$.
In group $B$,there are $6$ science and $7$ engineering books,total $13$ books. So,$P(E|B) = \frac{6}{13}$.
Using the law of total probability:
$P(E) = P(A) \times P(E|A) + P(B) \times P(E|B)$
$P(E) = \frac{1}{3} \times \frac{8}{13} + \frac{2}{3} \times \frac{6}{13}$
$P(E) = \frac{8}{39} + \frac{12}{39} = \frac{20}{39}$

(A) Let $E_1$ be the event that $A$ and $B$ meet at a party,$E_2$ be the event that $A$ and $B$ meet at a sports club,and $D$ be the event that $A$ and $B$ dine together.
Given $P(E_2) = \frac{4}{9}$,so $P(E_1) = 1 - \frac{4}{9} = \frac{5}{9}$.
The conditional probabilities are $P(D|E_2) = \frac{2}{5}$ and $P(D|E_1) = \frac{1}{3}$.
Using the law of total probability,the probability that they dine together is:
$P(D) = P(E_1) \cdot P(D|E_1) + P(E_2) \cdot P(D|E_2)$
$P(D) = \frac{5}{9} \times \frac{1}{3} + \frac{4}{9} \times \frac{2}{5} = \frac{5}{27} + \frac{8}{45} = \frac{25 + 24}{135} = \frac{49}{135}$.
The probability that they disperse without dining together is $P(D') = 1 - P(D) = 1 - \frac{49}{135} = \frac{86}{135}$.

(B) The total number of cards in a deck is $52$,and the number of aces is $4$.
When the first card is drawn,the probability of getting an ace is $P(A_1) = \frac{4}{52} = \frac{1}{13}$.
Since the card is drawn without replacement,there are now $51$ cards left in the deck,and $3$ of them are aces.
The probability of drawing a second ace,given that the first card was an ace,is $P(A_2|A_1) = \frac{3}{51} = \frac{1}{17}$.
The probability that both cards are aces is $P(A_1 \cap A_2) = P(A_1) \times P(A_2|A_1)$.
$P(A_1 \cap A_2) = \frac{1}{13} \times \frac{1}{17} = \frac{1}{221}$.

(C) Two events $A$ and $B$ are independent if and only if $P(A \cap B) = P(A) \cdot P(B)$.
We check option $(C)$:
$P(A^C \cap B^C) = P((A \cup B)^C)$ (by De Morgan's Law)
$= 1 - P(A \cup B)$
$= 1 - [P(A) + P(B) - P(A \cap B)]$
Since $A$ and $B$ are independent,$P(A \cap B) = P(A) \cdot P(B)$.
$= 1 - P(A) - P(B) + P(A) \cdot P(B)$
$= (1 - P(A)) - P(B)(1 - P(A))$
$= (1 - P(A))(1 - P(B))$
Thus,if $A$ and $B$ are independent,then $P(A^C \cap B^C) = (1 - P(A))(1 - P(B))$.

(B) The probability of getting $k$ successes in $n$ trials is given by the binomial distribution formula $P(X=r) = {}^{n}C_r p^r q^{n-r}$.
Here,$n=15$,$p=1/2$ (probability of tail),and $q=1/2$ (probability of head).
We need to find the probability of getting at least $3$ tails,i.e.,$P(X \geq 3)$.
This can be calculated as $P(X \geq 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]$.
$P(X=0) = {}^{15}C_0 (1/2)^0 (1/2)^{15} = 1 \times (1/2)^{15} = 1/2^{15}$.
$P(X=1) = {}^{15}C_1 (1/2)^1 (1/2)^{14} = 15 \times (1/2)^{15} = 15/2^{15}$.
$P(X=2) = {}^{15}C_2 (1/2)^2 (1/2)^{13} = \frac{15 \times 14}{2} \times (1/2)^{15} = 105/2^{15}$.
Summing these probabilities: $P(X < 3) = \frac{1 + 15 + 105}{2^{15}} = \frac{121}{2^{15}}$.
Therefore,$P(X \geq 3) = 1 - \frac{121}{2^{15}}$.

(A) The problem follows a Poisson distribution because the number of trials $n = 2000$ is large and the probability of success $p = 0.001$ is very small.
For a Poisson distribution,the parameter $\lambda$ is given by $\lambda = n \times p$.
$\lambda = 2000 \times 0.001 = 2$.
The probability mass function for a Poisson distribution is $P(X = x) = \frac{e^{-\lambda} \lambda^{x}}{x!}$.
We need to find the probability for exactly $x = 3$ individuals.
Substituting the values,we get $P(X = 3) = \frac{e^{-2} \times 2^{3}}{3!}$.
$P(X = 3) = \frac{e^{-2} \times 8}{6} = \frac{4}{3 e^{2}}$.

(C) For a Poisson distribution with mean '$m$',the probability mass function is given by $P(x = k) = \frac{m^k \cdot e^{-m}}{k!}$ for $k = 0, 1, 2, \dots$.
We need to find $P(x > 0)$.
Using the complement rule,$P(x > 0) = 1 - P(x = 0)$.
Substituting $k = 0$ into the formula,we get $P(x = 0) = \frac{m^0 \cdot e^{-m}}{0!} = \frac{1 \cdot e^{-m}}{1} = e^{-m}$.
Therefore,$P(x > 0) = 1 - e^{-m} = 1 - \frac{1}{e^m}$.
Simplifying this expression,we get $P(x > 0) = \frac{e^m - 1}{e^m}$.

(C) The sum of all probabilities for a random variable must be $1$.
$P(X = 0) + P(X = 1) + P(X = 2) = 1$
$3c^3 + (4c - 10c^2) + (5c - 1) = 1$
$3c^3 - 10c^2 + 9c - 2 = 0$
Factoring the cubic equation,we get $(c - 1)(c - 2)(3c - 1) = 0$.
Since $0 \le P(X) \le 1$,we test the values. If $c = 1$,$P(X = 2) = 5(1) - 1 = 4$,which is impossible. If $c = 2$,$P(X = 2) = 5(2) - 1 = 9$,which is impossible. Thus,$c = \frac{1}{3}$.
Now,$P(X = 0) = 3(\frac{1}{3})^3 = \frac{1}{9}$,$P(X = 1) = 4(\frac{1}{3}) - 10(\frac{1}{3})^2 = \frac{4}{3} - \frac{10}{9} = \frac{2}{9}$,and $P(X = 2) = 5(\frac{1}{3}) - 1 = \frac{2}{3}$.
We need to find $P(0 < X \le 2) = P(X = 1) + P(X = 2)$.
$P(0 < X \le 2) = \frac{2}{9} + \frac{2}{3} = \frac{2}{9} + \frac{6}{9} = \frac{8}{9}$.

(A) For a discrete probability distribution,the sum of all probabilities must be equal to $1$.
$\sum_{x=1}^{\infty} P(X = x) = 1$
$\sum_{x=1}^{\infty} 5r^x = 1$
$5(r + r^2 + r^3 + \dots) = 1$
This is an infinite geometric series with the first term $a = r$ and common ratio $r$. The sum is given by $\frac{a}{1-r}$ for $|r| < 1$.
$5 \left( \frac{r}{1 - r} \right) = 1$
$5r = 1 - r$
$6r = 1$
$r = \frac{1}{6}$

(D) For any probability distribution,the sum of all probabilities must be equal to $1$.
$\sum_{x=0}^{7} P(x) = 1$
$\Rightarrow 0.01 + 0.10 + 0.26 + 0.33 + 0.18 + 0.06 + K + 0.04 = 1$
$\Rightarrow 0.98 + K = 1$
$\Rightarrow K = 1 - 0.98 = 0.02$
Now,we need to calculate $P(X \geq 3) - P(X < 6)$.
$P(X \geq 3) = P(3) + P(4) + P(5) + P(6) + P(7) = 0.33 + 0.18 + 0.06 + 0.02 + 0.04 = 0.63$
$P(X < 6) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) = 0.01 + 0.10 + 0.26 + 0.33 + 0.18 + 0.06 = 0.94$
Therefore,$P(X \geq 3) - P(X < 6) = 0.63 - 0.94 = -0.31$.

(A) The probability of a mask being defective is $p = \frac{1}{500} = 0.002$.
In a packet of $n = 100$ masks,the expected number of defective masks is $\lambda = np = 100 \times 0.002 = 0.2$.
Using Poisson distribution,the probability that a packet contains $r$ defective masks is $P(X = r) = \frac{e^{-\lambda} \lambda^r}{r!}$.
For a packet to contain no defective masks,we set $r = 0$:
$P(X = 0) = \frac{e^{-0.2} (0.2)^0}{0!} = e^{-0.2}$.
In a consignment of $10,000$ packets,the number of packets with no defective masks is $10,000 \times P(X = 0) = 10,000 \times e^{-0.2} = \frac{10,000}{e^{0.2}}$.

(D) The distribution of successes follows a binomial distribution with parameters $n$ and $p$.
The variance of a binomial distribution is given by the formula $Var(X) = npq$,where $n$ is the number of trials,$p$ is the probability of success,and $q = 1 - p$ is the probability of failure.
In a single roll of a die,the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$. The odd numbers are $\{1, 3, 5\}$.
Therefore,the probability of success $p = \frac{3}{6} = \frac{1}{2}$.
The probability of failure $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.
Given that the die is rolled $n = 5$ times.
Substituting these values into the variance formula:
$Var(X) = n \times p \times q = 5 \times \frac{1}{2} \times \frac{1}{2} = \frac{5}{4}$.

(D) Given the probability function $P(X=r) = K r^2$ for $r \in \{-2, -1, 0, 1, 2, 3\}$.
Since the sum of all probabilities must be $1$,we have:
$\sum P(X=r) = 1$
$K((-2)^2 + (-1)^2 + 0^2 + 1^2 + 2^2 + 3^2) = 1$
$K(4 + 1 + 0 + 1 + 4 + 9) = 1$
$19K = 1 \Rightarrow K = \frac{1}{19}$
We need to find the sum of the variance $\sigma^2$ and the square of the mean $\mu^2$. We know that $\sigma^2 = E(X^2) - \mu^2$,so $\sigma^2 + \mu^2 = E(X^2)$.
$E(X^2) = \sum r^2 P(X=r) = \sum r^2 (K r^2) = K \sum r^4$
$E(X^2) = K((-2)^4 + (-1)^4 + 0^4 + 1^4 + 2^4 + 3^4)$
$E(X^2) = K(16 + 1 + 0 + 1 + 16 + 81) = K(115)$
Substituting $K = \frac{1}{19}$,we get:
$E(X^2) = \frac{115}{19}$
Thus,$\sigma^2 + \mu^2 = \frac{115}{19}$.

(B) For a Poisson distribution,the probability mass function is given by $P(X = x) = \frac{e^{-\lambda} \cdot \lambda^x}{x!}$.
Given that $P(X = 2) = 2 \cdot P(X = 1)$.
Substituting the formula:
$\frac{e^{-\lambda} \cdot \lambda^2}{2!} = 2 \cdot \frac{e^{-\lambda} \cdot \lambda^1}{1!}$.
Dividing both sides by $e^{-\lambda} \cdot \lambda$ (assuming $\lambda \neq 0$):
$\frac{\lambda}{2} = 2 \cdot 1$.
$\lambda = 4$.
In a Poisson distribution,the variance is equal to the parameter $\lambda$,so $\sigma^2 = \lambda = 4$.
The standard deviation $\sigma$ is $\sqrt{\lambda} = \sqrt{4} = 2$.

(C) We know that the sum of all probabilities in a distribution must be equal to $1$.
$\sum P(X = x) = 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k = 1$
$10k^2 + 9k - 1 = 0$
$(10k - 1)(k + 1) = 0$
Since $k$ must be positive for probabilities to be valid,we have $k = \frac{1}{10}$.
We need to find $P(0 < X < 4) = P(X = 1) + P(X = 2) + P(X = 3)$.
$P(0 < X < 4) = k + 2k + 2k = 5k$.
Substituting $k = \frac{1}{10}$,we get $P(0 < X < 4) = 5 \times \frac{1}{10} = \frac{1}{2}$.