AP EAMCET 2021 Mathematics Question Paper with Answer and Solution

(A) The equation of the family of circles passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + \lambda S_2 = 0$.
Given $S_1: x^2+y^2-8x-6y+21=0$ and $S_2: x^2+y^2-2x-15=0$.
The equation is $(x^2+y^2-8x-6y+21) + \lambda(x^2+y^2-2x-15) = 0$.
Since the circle passes through $(1,2)$,substitute $x=1$ and $y=2$:
$(1^2+2^2-8(1)-6(2)+21) + \lambda(1^2+2^2-2(1)-15) = 0$
$(1+4-8-12+21) + \lambda(5-2-15) = 0$
$6 + \lambda(-12) = 0 \Rightarrow \lambda = \frac{1}{2}$.
Substituting $\lambda = \frac{1}{2}$ into the family equation:
$(x^2+y^2-8x-6y+21) + \frac{1}{2}(x^2+y^2-2x-15) = 0$
Multiply by $2$:
$2(x^2+y^2-8x-6y+21) + (x^2+y^2-2x-15) = 0$
$3x^2+3y^2-18x-12y+27 = 0$
Dividing by $3$:
$x^2+y^2-6x-4y+9 = 0$.

(B) The circumference of the circle is $4 \pi$.
Since the circumference is given by $2 \pi r = 4 \pi$,we have $r = 2$.
The equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.
Substituting the center $(-2, 3)$ and radius $r = 2$:
$(x - (-2))^2 + (y - 3)^2 = 2^2$
$(x + 2)^2 + (y - 3)^2 = 4$
$x^2 + 4x + 4 + y^2 - 6y + 9 = 4$
$x^2 + y^2 + 4x - 6y + 9 = 0$.

(C) The given equation of the circle is $x^2+y^2-6x+4y-12=0$.
Completing the square,we get $(x-3)^2+(y+2)^2=5^2$.
The center is $(h, k) = (3, -2)$ and the radius $r=5$.
The parametric coordinates are $x=3+5\cos\theta$ and $y=-2+5\sin\theta$.
For point $A$ with $\theta=30^{\circ}$,$A = (3+5\cos 30^{\circ}, -2+5\sin 30^{\circ}) = (3+\frac{5\sqrt{3}}{2}, -2+\frac{5}{2}) = (\frac{6+5\sqrt{3}}{2}, \frac{1}{2})$.
For point $B$ with $\theta=90^{\circ}$,$B = (3+5\cos 90^{\circ}, -2+5\sin 90^{\circ}) = (3, 3)$.
The slope of chord $AB$ is $m = \frac{3-1/2}{3-(6+5\sqrt{3})/2} = \frac{5/2}{(6-5\sqrt{3})/2} = \frac{5}{6-5\sqrt{3}}$.
Wait,calculating slope $m = \frac{3-0.5}{3-(3+2.5\sqrt{3})} = \frac{2.5}{-2.5\sqrt{3}} = -\frac{1}{\sqrt{3}}$.
Using the point-slope form $y-y_1 = m(x-x_1)$ at point $B(3, 3)$:
$y-3 = -\frac{1}{\sqrt{3}}(x-3)$.
$\sqrt{3}y-3\sqrt{3} = -x+3$.
$x+\sqrt{3}y-3-3\sqrt{3} = 0$.
$x+\sqrt{3}y-3(1+\sqrt{3}) = 0$.

(B) The center of the circle is the point of intersection of its diameters.
Given equations of diameters:
$x + 2y - 5 = 0$ ... $(i)$
$3x - y - 1 = 0$ ... $(ii)$
Multiplying $(ii)$ by $2$,we get $6x - 2y - 2 = 0$ ... $(iii)$
Adding $(i)$ and $(iii)$:
$(x + 2y - 5) + (6x - 2y - 2) = 0$
$7x - 7 = 0 \Rightarrow x = 1$
Substituting $x = 1$ in $(i)$:
$1 + 2y - 5 = 0$ $\Rightarrow 2y = 4$ $\Rightarrow y = 2$
So,the center $(h, k) = (1, 2)$ and the radius $r = 5$.
The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$.
$(x - 1)^2 + (y - 2)^2 = 5^2$
$x^2 - 2x + 1 + y^2 - 4y + 4 = 25$
$x^2 + y^2 - 2x - 4y + 5 - 25 = 0$
$x^2 + y^2 - 2x - 4y - 20 = 0$

(D) Since the circle touches the $X$-axis at $(1, 0)$,the $x$-coordinate of the center is $1$.
Since the circle touches the $Y$-axis at $(0, 1)$,the $y$-coordinate of the center is $1$.
Thus,the center of the circle is $(1, 1)$ and the radius is $r = 1$.
The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.
Substituting the values,we get $(x-1)^2 + (y-1)^2 = 1^2$.
Expanding this,we have $(x^2 - 2x + 1) + (y^2 - 2y + 1) = 1$.
Simplifying,we get $x^2 + y^2 - 2x - 2y + 1 = 0$.

(C) The power of a point $(x_1, y_1)$ with respect to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by the expression $x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$.
Given the point $(1, 6)$ and the circle $x^2 + y^2 + 4x - 6y - a = 0$,the power is:
$1^2 + 6^2 + 4(1) - 6(6) - a = -16$
$1 + 36 + 4 - 36 - a = -16$
$5 - a = -16$
$a = 5 + 16$
$a = 21$

(C) For a point $(x_1, y_1)$ to lie inside the circle $x^2+y^2-r^2=0$,the condition is $x_1^2+y_1^2-r^2 < 0$.
Given the point $(\lambda, 1+\lambda)$ and the circle $x^2+y^2-1=0$,we substitute the point into the equation:
$\lambda^2 + (1+\lambda)^2 - 1 < 0$
$\lambda^2 + 1 + 2\lambda + \lambda^2 - 1 < 0$
$2\lambda^2 + 2\lambda < 0$
$2\lambda(\lambda+1) < 0$
Dividing by $2$,we get $\lambda(\lambda+1) < 0$.
This inequality holds when $\lambda$ lies between the roots $-1$ and $0$.
Therefore,$-1 < \lambda < 0$.

(D) The equation of the circle is $x^2+y^2-4x-8y-5=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-2, f=-4, c=-5$.
The center of the circle is $(2, 4)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-2)^2+(-4)^2-(-5)} = \sqrt{4+16+5} = \sqrt{25} = 5$.
For the line $3x-4y-m=0$ to intersect the circle at two distinct points,the perpendicular distance $d$ from the center $(2, 4)$ to the line must be less than the radius $r$.
$d = \frac{|3(2)-4(4)-m|}{\sqrt{3^2+(-4)^2}} = \frac{|6-16-m|}{\sqrt{9+16}} = \frac{|-10-m|}{5} = \frac{|m+10|}{5}$.
Setting $d < r$,we have $\frac{|m+10|}{5} < 5$,which implies $|m+10| < 25$.
This inequality is equivalent to $-25 < m+10 < 25$.
Subtracting $10$ from all parts,we get $-35 < m < 15$.
The integers $m$ satisfying this are $\{-34, -33, \dots, 14\}$.
The number of such integers is $14 - (-34) + 1 = 14 + 34 + 1 = 49$.

(A) Given the circle equation: $x^2+y^2-6x-4y-12=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3, f=-2, c=-12$.
Centre of the circle: $(-g, -f) = (3, 2)$.
Radius of the circle: $r = \sqrt{g^2+f^2-c} = \sqrt{(-3)^2+(-2)^2-(-12)} = \sqrt{9+4+12} = \sqrt{25} = 5$.
Since the lines are parallel to the $x$-axis,they are of the form $y = k$.
These lines touch the circle,so the distance from the centre $(3, 2)$ to the line $y = k$ must be equal to the radius $r = 5$.
Therefore,$|k - 2| = 5$.
This gives $k - 2 = 5$ or $k - 2 = -5$.
So,$k = 7$ or $k = -3$.
The equations of the lines are $y = 7$ and $y = -3$.
The pair of straight lines is given by $(y - 7)(y + 3) = 0$.
Expanding this,we get $y^2 + 3y - 7y - 21 = 0$,which simplifies to $y^2 - 4y - 21 = 0$.

(B) The given equation of the circle is $x^2 + y^2 + 2gx + 2fy + d = 0$.
Its centre is $(-g, -f)$.
$A$ normal to a circle always passes through its centre.
The given equation of the line is $ax + by + c = 0$.
Since the line is a normal,it must pass through the centre $(-g, -f)$.
Substituting the centre $(-g, -f)$ into the line equation:
$a(-g) + b(-f) + c = 0$
$-ag - bf + c = 0$
Multiplying by $-1$,we get:
$ag + bf - c = 0$

(C) Let the radius of the circle be $r$. Since the circle lies in the fourth quadrant and touches the lines $x=0$ and $y=0$,its center must be at $(r, -r)$,where $r > 0$.
The distance from the center $(r, -r)$ to the line $3x+4y-12=0$ must be equal to the radius $r$.
Using the perpendicular distance formula,we have:
$\left| \frac{3(r) + 4(-r) - 12}{\sqrt{3^2 + 4^2}} \right| = r$
$\left| \frac{3r - 4r - 12}{5} \right| = r$
$\left| \frac{-r - 12}{5} \right| = r$
$|r + 12| = 5r$
Since $r > 0$,$r + 12 = 5r$ or $r + 12 = -5r$.
Case $1$: $4r = 12 \Rightarrow r = 3$.
Case $2$: $6r = -12 \Rightarrow r = -2$ (rejected as $r > 0$).
Thus,the radius is $3$ units.

(C) The given equation of the circle is $x^2+y^2-8x-2y-8=0$.
Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=-4$,$f=-1$,and $c=-8$.
The centre of the circle is $(-g, -f) = (4, 1)$ and the radius $r$ is $\sqrt{g^2+f^2-c} = \sqrt{16+1+8} = \sqrt{25} = 5$.
The perpendicular distance $d$ from the centre $(4, 1)$ to the line $x+y+1=0$ is given by:
$d = \left|\frac{4+1+1}{\sqrt{1^2+1^2}}\right| = \frac{6}{\sqrt{2}} = 3\sqrt{2}$.
The length of the chord is given by $2\sqrt{r^2-d^2}$.
Length $= 2\sqrt{5^2 - (3\sqrt{2})^2} = 2\sqrt{25 - 18} = 2\sqrt{7}$ units.

(A) The equation of the circle is $x^2+y^2=16$,which represents a circle centered at the origin $(0,0)$ with radius $r=4$.
Let the two points be $A = (4 \cos \theta, 4 \sin \theta)$ and $B = (4 \cos (\theta+60^{\circ}), 4 \sin (\theta+60^{\circ}))$.
These points represent points on the circle with polar angles $\theta$ and $\theta+60^{\circ}$ respectively.
The angle subtended by the chord $AB$ at the center $O(0,0)$ is $\Delta \phi = (\theta+60^{\circ}) - \theta = 60^{\circ}$.
Since $OA = OB = r = 4$ and the included angle $\angle AOB = 60^{\circ}$,the triangle $\triangle OAB$ is an equilateral triangle.
Therefore,the length of the chord $AB$ is equal to the radius of the circle.
$AB = r = 4$.

(A) line passing through the point $(4,0)$ with slope $m$ is given by $y - 0 = m(x - 4)$,which simplifies to $mx - y - 4m = 0$.
For this line to be a tangent to the circle $x^2 + y^2 = 4$ (with center $(0,0)$ and radius $r = 2$),the perpendicular distance from the center to the line must equal the radius.
Using the distance formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$,we have $\frac{|m(0) - (0) - 4m|}{\sqrt{m^2 + (-1)^2}} = 2$.
$| -4m | = 2\sqrt{m^2 + 1}$.
Squaring both sides: $16m^2 = 4(m^2 + 1)$ $\Rightarrow 4m^2 = m^2 + 1$ $\Rightarrow 3m^2 = 1$ $\Rightarrow m = \pm \frac{1}{\sqrt{3}}$.
Substituting $m$ back into the line equation: $y = \pm \frac{1}{\sqrt{3}}(x - 4)$.

(B) The given circle is $x^2+y^2+4x+8y-38=0$.
Completing the square,we get $(x+2)^2+(y+4)^2=58$.
The center of the circle is $C(-2,-4)$.
Let $Q(x_1, y_1)$ be the other end of the diameter passing through $P(-9,-1)$.
Since $C$ is the midpoint of $PQ$,we have $\frac{x_1-9}{2}=-2 \Rightarrow x_1=5$ and $\frac{y_1-1}{2}=-4 \Rightarrow y_1=-7$.
Thus,$Q$ is $(5,-7)$.
The tangent at $Q$ is parallel to the tangent at $P$.
The slope of the radius $CP$ is $m_{CP} = \frac{-1-(-4)}{-9-(-2)} = \frac{3}{-7} = -\frac{3}{7}$.
The slope of the tangent at $P$ (and $Q$) is the negative reciprocal of the slope of the radius,which is $m = -\frac{1}{-3/7} = \frac{7}{3}$.
The equation of the tangent at $Q(5,-7)$ is $y - (-7) = \frac{7}{3}(x - 5)$.
$3(y+7) = 7(x-5)$ $\Rightarrow 3y+21 = 7x-35$ $\Rightarrow 7x-3y=56$.

(C) For a point $A$ outside a circle,if a secant through $A$ intersects the circle at points $C$ and $B$,then the length of the tangent $AT$ from $A$ to the circle is given by the Power of a Point theorem: $AT^2 = AC \cdot AB$.
First,calculate the distances $AC$ and $AB$ using the distance formula:
$AC = \sqrt{(5-3)^2 + (0-1)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4+1} = \sqrt{5}$
$AB = \sqrt{(7-3)^2 + (-1-1)^2} = \sqrt{4^2 + (-2)^2} = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5}$
Now,substitute these values into the formula:
$AT^2 = AC \cdot AB = \sqrt{5} \cdot 2\sqrt{5} = 2 \cdot 5 = 10$
$AT = \sqrt{10}$ units.

(C) Let the circle pass through points $A(2,2)$ and $B(9,9)$ and touch the $X$-axis at point $P$.
By the power of a point theorem,since $OP$ is a tangent to the circle from the origin $O$ and $OAB$ is a secant line,we have:
$OP^2 = OA \cdot OB$
Calculating the distances $OA$ and $OB$ from the origin $O(0,0)$:
$OA = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$
$OB = \sqrt{9^2 + 9^2} = \sqrt{81 + 81} = \sqrt{162} = 9\sqrt{2}$
Substituting these values into the equation:
$OP^2 = (2\sqrt{2}) \cdot (9\sqrt{2}) = 18 \cdot 2 = 36$
$OP = \sqrt{36} = 6$
Thus,$OP = 6$.

(D) The equation of the circle is $S: x^2 + y^2 + 8x - 6y + k = 0$.
The length of the tangent drawn from an external point $(x_1, y_1)$ to the circle $S = 0$ is given by $\sqrt{S_1}$,where $S_1 = x_1^2 + y_1^2 + 8x_1 - 6y_1 + k$.
Given the point is $(-2, 3)$ and the length of the tangent is $4$ units:
$4 = \sqrt{(-2)^2 + (3)^2 + 8(-2) - 6(3) + k}$
$4 = \sqrt{4 + 9 - 16 - 18 + k}$
$4 = \sqrt{k - 21}$
Squaring both sides:
$16 = k - 21$
$k = 16 + 21 = 37$.

(D) The length of the tangent from a point $(x_1, y_1)$ to a circle $S: x^2+y^2+2gx+2fy+c=0$ is given by $\sqrt{S_1} = \sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c}$.
For the first circle $C_1: x^2+y^2+x+y-4=0$,the length of the tangent $L_1$ from $(1,2)$ is:
$L_1 = \sqrt{1^2+2^2+1+2-4} = \sqrt{1+4+1+2-4} = \sqrt{4} = 2$.
For the second circle $C_2: 3x^2+3y^2-x-y-k=0$,we first normalize it to $x^2+y^2-\frac{1}{3}x-\frac{1}{3}y-\frac{k}{3}=0$.
The length of the tangent $L_2$ from $(1,2)$ is:
$L_2 = \sqrt{1^2+2^2-\frac{1}{3}(1)-\frac{1}{3}(2)-\frac{k}{3}} = \sqrt{1+4-\frac{1}{3}-\frac{2}{3}-\frac{k}{3}} = \sqrt{5-1-\frac{k}{3}} = \sqrt{4-\frac{k}{3}}$.
Given the ratio $\frac{L_1}{L_2} = \frac{4}{3}$,we have:
$\frac{2}{\sqrt{4-\frac{k}{3}}} = \frac{4}{3} \Rightarrow \frac{1}{\sqrt{4-\frac{k}{3}}} = \frac{2}{3}$.
Squaring both sides:
$\frac{1}{4-\frac{k}{3}} = \frac{4}{9} \Rightarrow 4-\frac{k}{3} = \frac{9}{4}$.
$\frac{k}{3} = 4 - \frac{9}{4} = \frac{16-9}{4} = \frac{7}{4}$.
$k = 3 \times \frac{7}{4} = \frac{21}{4}$.

(B) Given,the equation of the circle is $5x^2 + 5y^2 = 1$.
Dividing by $5$,we get $x^2 + y^2 = \frac{1}{5} = \left(\frac{1}{\sqrt{5}}\right)^2$.
Thus,the center is $(0, 0)$ and the radius $r = \frac{1}{\sqrt{5}}$.
The given line is $3x + 4y = 1$,which has a slope $m = -\frac{3}{4}$.
The equation of a tangent to the circle $x^2 + y^2 = r^2$ with slope $m$ is given by $y = mx \pm r\sqrt{1 + m^2}$.
Substituting $m = -\frac{3}{4}$ and $r = \frac{1}{\sqrt{5}}$:
$y = -\frac{3}{4}x \pm \frac{1}{\sqrt{5}} \sqrt{1 + \left(-\frac{3}{4}\right)^2}$
$y = -\frac{3}{4}x \pm \frac{1}{\sqrt{5}} \sqrt{1 + \frac{9}{16}}$
$y = -\frac{3}{4}x \pm \frac{1}{\sqrt{5}} \sqrt{\frac{25}{16}}$
$y = -\frac{3}{4}x \pm \frac{1}{\sqrt{5}} \cdot \frac{5}{4}$
$y = -\frac{3}{4}x \pm \frac{\sqrt{5}}{4}$
Multiplying by $4$:
$4y = -3x \pm \sqrt{5}$
$3x + 4y = \pm \sqrt{5}$.

(A) Given the circle equation $x^2+y^2=50$ and the line $x+7=0$.
Substituting $x = -7$ into the circle equation:
$(-7)^2 + y^2 = 50$
$49 + y^2 = 50$
$y^2 = 1 \Rightarrow y = \pm 1$.
Thus,the points of intersection are $P_1(-7, 1)$ and $P_2(-7, -1)$.
The equation of the tangent to the circle $x^2+y^2=r^2$ at $(x_1, y_1)$ is given by $xx_1 + yy_1 = r^2$.
For point $(-7, 1)$: $-7x + y = 50 \Rightarrow 7x - y + 50 = 0$.
For point $(-7, -1)$: $-7x - y = 50 \Rightarrow 7x + y + 50 = 0$.
Therefore,the required equations are $7x+y+50=0$ and $7x-y+50=0$.

(D) Let point $P(x_1, y_1)$ be any point on the circle $(x-3)^2+(y+2)^2=5r^2$.
Since $P$ lies on this circle, it satisfies the equation:
$(x_1-3)^2+(y_1+2)^2=5r^2 \dots (i)$
The length of the tangent drawn from point $P(x_1, y_1)$ to the circle $(x-3)^2+(y+2)^2=r^2$ is given by $\sqrt{S_1}$, where $S_1 = (x_1-3)^2+(y_1+2)^2-r^2$.
Substituting $(i)$ into the expression:
Length $= \sqrt{5r^2-r^2} = \sqrt{4r^2} = 2r$.
Given that the length of the tangent is $16$ units, we have $2r = 16$, which implies $r = 8$.
The area between the two concentric circles is the difference of their areas:
Area $= \pi(R^2) - \pi(r^2) = \pi(5r^2) - \pi(r^2) = 4\pi r^2$.
Substituting $r = 8$:
Area $= 4 \pi (8)^2 = 4 \pi (64) = 256 \pi$ sq. units.

(A) The given equation of the circle is $x^2+y^2+4x+4y-1=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=2, f=2, c=-1$.
The center of the circle is $O(-g, -f) = (-2, -2)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{4+4-(-1)} = \sqrt{9} = 3$.
Let the external point be $C(1, 1)$. The distance $OC$ is $\sqrt{(1 - (-2))^2 + (1 - (-2))^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$.
Let $\alpha$ be the angle between the line joining the center to the external point and the tangent. In the right-angled triangle $\triangle OAC$ (where $A$ is the point of tangency),$\sin \alpha = \frac{OA}{OC} = \frac{r}{OC} = \frac{3}{3\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Thus,$\alpha = 45^\circ$ or $\frac{\pi}{4}$.
The angle between the pair of tangents is $2\alpha = 2 \times \frac{\pi}{4} = \frac{\pi}{2}$.

(A) Let the tangent to the circle $x^2+y^2=4$ be $x \cos \theta + y \sin \theta = 2$.
This line can be written as $x \cos \theta + y \sin \theta - 2 = 0$.
Let $(x_1, y_1)$ be the pole of this tangent with respect to the circle $(x+2)^2+y^2=8$,which expands to $x^2+4x+4+y^2=8$ or $x^2+y^2+4x-4=0$.
The equation of the polar of $(x_1, y_1)$ with respect to $x^2+y^2+4x-4=0$ is given by $x x_1 + y y_1 + 2(x+x_1) - 4 = 0$.
Rearranging this,we get $(x_1+2)x + y_1 y + (2x_1-4) = 0$.
Since this represents the same line as the tangent,we compare the coefficients:
$\frac{\cos \theta}{x_1+2} = \frac{\sin \theta}{y_1} = \frac{-2}{2x_1-4} = \frac{-1}{x_1-2}$.
Thus,$\cos \theta = -\frac{x_1+2}{x_1-2}$ and $\sin \theta = -\frac{y_1}{x_1-2}$.
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$,we have $\left(-\frac{x_1+2}{x_1-2}\right)^2 + \left(-\frac{y_1}{x_1-2}\right)^2 = 1$.
This simplifies to $(x_1+2)^2 + y_1^2 = (x_1-2)^2$.
$x_1^2 + 4x_1 + 4 + y_1^2 = x_1^2 - 4x_1 + 4$.
$y_1^2 + 8x_1 = 0$.
Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $y^2+8x=0$.

(C) The given equation of the circle is $x^2+y^2-14x+2y+25=0$.
Completing the square,we get $(x^2-14x+49) + (y^2+2y+1) - 49 - 1 + 25 = 0$,which simplifies to $(x-7)^2 + (y+1)^2 = 25 = 5^2$.
Thus,the radius $r = 5$ and the center $P = (7, -1)$.
The distance from the origin $O(0,0)$ to the center $P(7,-1)$ is $OP = \sqrt{7^2 + (-1)^2} = \sqrt{49+1} = \sqrt{50} = 5\sqrt{2}$.
Let the tangents from the origin touch the circle at $A$ and $B$. In the right-angled triangle $\triangle OAP$,$\sin \theta = \frac{AP}{OP} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Therefore,$\theta = 45^{\circ}$.
Similarly,for $\triangle OBP$,$\sin \alpha = \frac{BP}{OP} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$,so $\alpha = 45^{\circ}$.
The total angle between the tangents is $\theta + \alpha = 45^{\circ} + 45^{\circ} = 90^{\circ}$.

(C) Let $C_1: x^2+y^2=r_1^2$,$C_2: x^2+y^2=r_2^2$,and $C_3: x^2+y^2=r_3^2$.
Let $P(x_1, y_1)$ be a point on the circle $C_1$,so $x_1^2+y_1^2=r_1^2$.
The equation of the chord of contact of tangents from $P$ to the circle $C_2$ is given by $T=0$,which is $x x_1+y y_1-r_2^2=0$.
Since this line touches the circle $C_3$,the perpendicular distance from the origin $(0,0)$ to the line must be equal to the radius $r_3$.
Thus,$\frac{|0 \cdot x_1+0 \cdot y_1-r_2^2|}{\sqrt{x_1^2+y_1^2}}=r_3$.
Substituting $x_1^2+y_1^2=r_1^2$,we get $\frac{r_2^2}{\sqrt{r_1^2}}=r_3$,which simplifies to $r_2^2=r_1 r_3$.
Therefore,$r_1, r_2, r_3$ are in $GP$.

(D) The given circle equation is $x^2 + y^2 + 4x + 16y - 30 = 0$.
Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g = 4$,$2f = 16$,and $c = -30$.
Thus,$g = 2$,$f = 8$,and $c = -30$.
The centre of this circle is $C_1 = (-g, -f) = (-2, -8)$.
The radius of this circle is $r_1 = \sqrt{g^2 + f^2 - c} = \sqrt{2^2 + 8^2 - (-30)} = \sqrt{4 + 64 + 30} = \sqrt{98}$.
Let the other circle have centre $C_2 = (1, 2)$ and radius $r_2$.
The distance between the centres $d$ is given by $d^2 = (1 - (-2))^2 + (2 - (-8))^2 = 3^2 + 10^2 = 9 + 100 = 109$.
Since the two circles are orthogonal,they satisfy the condition $d^2 = r_1^2 + r_2^2$.
Substituting the values,$109 = 98 + r_2^2$.
$r_2^2 = 109 - 98 = 11$.
Therefore,$r_2 = \sqrt{11}$.

(A) Let the given circle be $S_1: x^2+y^2-8x-6y+23=0$. The center of $S_1$ is $(4, 3)$.
For a circle $S_2$ to bisect the circumference of $S_1$,the radical axis of $S_1$ and $S_2$ must pass through the center of $S_1$.
The radical axis is given by $S_1 - S_2 = 0$.
For option $A$,$S_2: x^2+y^2-6x-4y+9=0$.
The radical axis is $(x^2+y^2-8x-6y+23) - (x^2+y^2-6x-4y+9) = 0$.
This simplifies to $-2x - 2y + 14 = 0$,or $x + y - 7 = 0$.
Substituting the center $(4, 3)$ into the radical axis equation: $4 + 3 - 7 = 0$.
Since the center satisfies the equation,the circle $x^2+y^2-6x-4y+9=0$ bisects the circumference of $S_1$.

(B) Let the equation of the required circle be $(x-h)^2+(y-k)^2=r^2$. Since it touches the $Y$-axis,the radius $r = |h|$. Thus,the equation is $(x-h)^2+(y-k)^2=h^2$,which simplifies to $x^2+y^2-2hx-2ky+k^2=0$.
Since it passes through $(3,0)$,we have $(3-h)^2+(0-k)^2=h^2$,which simplifies to $9-6h+k^2=0$,or $k^2=6h-9$ ... $(i)$.
The circle $x^2+y^2-6x+4y-3=0$ has $g_2=-3, f_2=2, c_2=-3$. The required circle has $g_1=-h, f_1=-k, c_1=k^2$.
For orthogonal intersection,$2(g_1g_2+f_1f_2) = c_1+c_2$.
Substituting the values: $2((-h)(-3)+(-k)(2)) = k^2-3$,which gives $6h-4k = k^2-3$.
Substituting $k^2=6h-9$ from $(i)$ into this equation: $6h-4k = (6h-9)-3$,which simplifies to $-4k = -12$,so $k=3$.
Substituting $k=3$ into $(i)$: $9-6h+9=0$,so $6h=18$,$h=3$.
The center is $(3,3)$ and the radius is $3$.
The equation is $(x-3)^2+(y-3)^2=3^2$,which simplifies to $x^2+y^2-6x-6y+9=0$.

(B) Given curves are: $x^2+y^2=4x$ $(i)$ and $x^2+y^2=8$ $(ii)$.
To find the angle between the curves at $(2,2)$,we find the slopes of the tangents at this point.
For curve $(i)$,differentiating with respect to $x$: $2x + 2y \frac{dy}{dx} = 4 \Rightarrow \frac{dy}{dx} = \frac{2-x}{y}$.
At $(2,2)$,$m_1 = \frac{2-2}{2} = 0$.
For curve $(ii)$,differentiating with respect to $x$: $2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$.
At $(2,2)$,$m_2 = -\frac{2}{2} = -1$.
The angle $\theta$ between the curves is given by $\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$.
$\tan \theta = \left| \frac{-1 - 0}{1 + 0 \times (-1)} \right| = |-1| = 1$.
Thus,$\theta = \tan^{-1}(1) = \frac{\pi}{4}$.
Given $\theta = \sin^{-1}(a)$,so $\sin^{-1}(a) = \frac{\pi}{4}$.
Therefore,$a = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.

(B) The given equations of the circles are:
$C_1: x^2+y^2-6x-8y-12=0$
$C_2: x^2+y^2-4x+6y+k=0$
Comparing these with the general form $x^2+y^2+2gx+2fy+c=0$:
For $C_1$: $g_1 = -3, f_1 = -4, c_1 = -12$
For $C_2$: $g_2 = -2, f_2 = 3, c_2 = k$
Two circles are orthogonal if $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Substituting the values:
$2(-3)(-2) + 2(-4)(3) = -12 + k$
$12 - 24 = -12 + k$
$-12 = -12 + k$
$k = 0$

(A) The given circles are $C_1: x^2+y^2+2x+4y+1=0$ and $C_2: x^2+y^2-2x+6y-3=0$.
Comparing with the general form $x^2+y^2+2gx+2fy+c=0$:
For $C_1$: $g_1=1, f_1=2, c_1=1$. Centre $O_1=(-1, -2)$,radius $r_1=\sqrt{1^2+2^2-1}=2$.
For $C_2$: $g_2=-1, f_2=3, c_2=-3$. Centre $O_2=(1, -3)$,radius $r_2=\sqrt{(-1)^2+3^2-(-3)}=\sqrt{1+9+3}=\sqrt{13}$.
The distance $d$ between the centres $O_1(-1, -2)$ and $O_2(1, -3)$ is $d=\sqrt{(1-(-1))^2+(-3-(-2))^2}=\sqrt{2^2+(-1)^2}=\sqrt{5}$.
The angle $\theta$ between the circles is given by $\cos \theta = \left|\frac{d^2-r_1^2-r_2^2}{2r_1r_2}\right|$.
Substituting the values: $\cos \theta = \left|\frac{5-4-13}{2 \times 2 \times \sqrt{13}}\right| = \left|\frac{-12}{4\sqrt{13}}\right| = \frac{3}{\sqrt{13}}$.
Therefore,$\theta = \cos^{-1}\left(\frac{3}{\sqrt{13}}\right)$.

(C) The length of the tangent from a point $(x_1, y_1)$ to a circle $S: x^2+y^2+2gx+2fy+c=0$ is given by $\sqrt{S_1} = \sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c}$.
For the first circle $C_1: x^2+y^2+x+y-4=0$,the length of the tangent $L_1$ from $(1,2)$ is:
$L_1 = \sqrt{1^2+2^2+1+2-4} = \sqrt{1+4+1+2-4} = \sqrt{4} = 2$.
For the second circle $C_2: 3x^2+3y^2-x-y-\lambda=0$,we first normalize the equation by dividing by $3$:
$x^2+y^2-\frac{1}{3}x-\frac{1}{3}y-\frac{\lambda}{3}=0$.
The length of the tangent $L_2$ from $(1,2)$ is:
$L_2 = \sqrt{1^2+2^2-\frac{1}{3}(1)-\frac{1}{3}(2)-\frac{\lambda}{3}} = \sqrt{1+4-\frac{1}{3}-\frac{2}{3}-\frac{\lambda}{3}} = \sqrt{5-1-\frac{\lambda}{3}} = \sqrt{4-\frac{\lambda}{3}}$.
Given the ratio $\frac{L_1}{L_2} = \frac{3}{4}$,we have:
$\frac{2}{\sqrt{4-\frac{\lambda}{3}}} = \frac{3}{4} \Rightarrow \sqrt{4-\frac{\lambda}{3}} = \frac{8}{3}$.
Squaring both sides:
$4-\frac{\lambda}{3} = \frac{64}{9} \Rightarrow \frac{\lambda}{3} = 4-\frac{64}{9} = \frac{36-64}{9} = -\frac{28}{9}$.
Multiplying by $3$,we get $\lambda = -\frac{28}{3}$.

(D) Let $S \equiv x^2+y^2+3x+5y+4=0$ and $S' \equiv x^2+y^2+5x+3y+4=0$.
The equation of the common chord is $S-S'=0$.
$\Rightarrow (x^2+y^2+3x+5y+4) - (x^2+y^2+5x+3y+4) = 0$
$\Rightarrow -2x+2y=0$
$\Rightarrow x-y=0$.
The center of $S=0$ is $C\left(-\frac{3}{2}, -\frac{5}{2}\right)$.
The radius $r$ is $\sqrt{\left(\frac{3}{2}\right)^2 + \left(\frac{5}{2}\right)^2 - 4} = \sqrt{\frac{9}{4} + \frac{25}{4} - 4} = \sqrt{\frac{34}{4} - \frac{16}{4}} = \sqrt{\frac{18}{4}} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}}$.
Let $p$ be the length of the perpendicular from $C$ to the line $x-y=0$:
$p = \frac{|-\frac{3}{2} - (-\frac{5}{2})|}{\sqrt{1^2+(-1)^2}} = \frac{|1|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.
The length of the common chord $AB = 2\sqrt{r^2-p^2}$.
$AB = 2\sqrt{\frac{9}{2} - \frac{1}{2}} = 2\sqrt{\frac{8}{2}} = 2\sqrt{4} = 2 \times 2 = 4$.

(D) The given circle is $x^2 + y^2 - 2x - 6y + 6 = 0$.
Completing the square,we get $(x - 1)^2 + (y - 3)^2 = 4$.
The center of this circle is $(1, 3)$ and its radius is $2$.
Since the chord of the required circle is a diameter of this circle,the chord lies on the line passing through the center $(1, 3)$ and its endpoints are the points where the chord intersects the circle.
However,the problem states the chord is a diameter of the circle $x^2 + y^2 - 2x - 6y + 6 = 0$. Thus,the chord is the line segment connecting the endpoints of the diameter of the given circle.
The endpoints of the diameter of $x^2 + y^2 - 2x - 6y + 6 = 0$ are $(1, 1)$ and $(1, 5)$.
The required circle has center $C(2, 1)$ and passes through the points $(1, 1)$ and $(1, 5)$.
The radius $r$ is the distance from $(2, 1)$ to $(1, 1)$:
$r = \sqrt{(2 - 1)^2 + (1 - 1)^2} = \sqrt{1^2 + 0^2} = 1$.
Alternatively,checking the distance to $(1, 5)$:
$r = \sqrt{(2 - 1)^2 + (1 - 5)^2} = \sqrt{1^2 + (-4)^2} = \sqrt{17}$.
Re-evaluating the problem statement: If the chord is a diameter of the circle $x^2 + y^2 - 2x - 6y + 6 = 0$,the chord is the line segment between $(1, 1)$ and $(1, 5)$,which is the vertical line $x = 1$.
The distance from the center $(2, 1)$ to the line $x = 1$ is $d = |2 - 1| = 1$.
The radius $r$ of the circle is the distance from the center $(2, 1)$ to any point on the chord,e.g.,$(1, 1)$.
$r = \sqrt{(2 - 1)^2 + (1 - 1)^2} = 1$.

(D) The equation of the radical axis of the two circles is obtained by subtracting the two equations: $(x^2+y^2+ax+by+c) - (x^2+y^2+bx+ay+c) = 0$.
This simplifies to $(a-b)x + (b-a)y = 0$,which is $(a-b)(x-y) = 0$.
Assuming $a \neq b$,the radical axis is the line $x-y=0$,or $y=x$.
Substituting $y=x$ into the first circle equation: $x^2+x^2+ax+bx+c=0 \Rightarrow 2x^2+(a+b)x+c=0$.
Let the roots be $x_1$ and $x_2$. Then $x_1+x_2 = -\frac{a+b}{2}$ and $x_1x_2 = \frac{c}{2}$.
The points of intersection are $(x_1, x_1)$ and $(x_2, x_2)$.
The length of the common chord is the distance between these points: $\sqrt{(x_2-x_1)^2 + (x_2-x_1)^2} = \sqrt{2(x_2-x_1)^2} = \sqrt{2} |x_2-x_1|$.
Using $|x_2-x_1| = \sqrt{(x_1+x_2)^2 - 4x_1x_2} = \sqrt{\frac{(a+b)^2}{4} - 2c} = \frac{\sqrt{(a+b)^2-8c}}{2}$.
Thus,the length is $\sqrt{2} \cdot \frac{\sqrt{(a+b)^2-8c}}{2} = \frac{\sqrt{(a+b)^2-8c}}{\sqrt{2}}$.
Therefore,Statement-$I$ is false.
Statement-$II$ is a standard theorem in geometry of circles,which is true.
Hence,Statement-$I$ is false and Statement-$II$ is true.

(B) Given circles are $S_1: x^2+y^2-2x+4y-4=0$ and $S_2: x^2+y^2+4x-6y-3=0$.
The equation of the common chord is given by $S_1 - S_2 = 0$.
$(x^2+y^2-2x+4y-4) - (x^2+y^2+4x-6y-3) = 0$.
$-6x + 10y - 1 = 0$,which simplifies to $6x - 10y + 1 = 0$.
The perpendicular distance from the point $(x_1, y_1) = (1, 2)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Substituting the values: $d = \frac{|6(1) - 10(2) + 1|}{\sqrt{6^2 + (-10)^2}}$.
$d = \frac{|6 - 20 + 1|}{\sqrt{36 + 100}} = \frac{|-13|}{\sqrt{136}} = \frac{13}{\sqrt{136}}$ units.

(A) Let the required circle be $x^2+y^2+2gx+2fy+c=0 \dots(1)$.
Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ cut orthogonally if $2g_1g_2+2f_1f_2=c_1+c_2$.
Applying this condition to circle $(1)$ and the given circles:
For circle $x^2+y^2-2x+3y-7=0$: $2g(-1)+2f(3/2)=c-7 \Rightarrow -2g+3f-c=-7 \dots(2)$.
For circle $x^2+y^2+5x-5y+9=0$: $2g(5/2)+2f(-5/2)=c+9 \Rightarrow 5g-5f-c=9 \dots(3)$.
For circle $x^2+y^2+7x-9y+29=0$: $2g(7/2)+2f(-9/2)=c+29 \Rightarrow 7g-9f-c=29 \dots(4)$.
Subtracting $(2)$ from $(3)$: $(5g-5f-c) - (-2g+3f-c) = 9 - (-7) \Rightarrow 7g-8f=16 \dots(5)$.
Subtracting $(3)$ from $(4)$: $(7g-9f-c) - (5g-5f-c) = 29 - 9$ $\Rightarrow 2g-4f=20$ $\Rightarrow g-2f=10$ $\Rightarrow g=2f+10$.
Substituting $g$ into $(5)$: $7(2f+10)-8f=16$ $\Rightarrow 14f+70-8f=16$ $\Rightarrow 6f=-54$ $\Rightarrow f=-9$.
Then $g=2(-9)+10=-8$.
From $(2)$: $-2(-8)+3(-9)-c=-7$ $\Rightarrow 16-27-c=-7$ $\Rightarrow -11-c=-7$ $\Rightarrow c=-4$.
Substituting $g, f, c$ in $(1)$: $x^2+y^2-16x-18y-4=0$.

(D) Given,the equation of the circle is $x^2+y^2+4x+6y-3=0$.
Given,the point is $P(1, 1)$.
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 $x \cdot x_1 + y \cdot y_1 + g(x+x_1) + f(y+y_1) + c = 0$.
Here,$g=2$,$f=3$,$c=-3$,$x_1=1$,and $y_1=1$.
Substituting these values into the formula:
$x(1) + y(1) + 2(x+1) + 3(y+1) - 3 = 0$
$x + y + 2x + 2 + 3y + 3 - 3 = 0$
$3x + 4y + 2 = 0$.

(B) The point having the same power with respect to three circles is their radical center.
To find the radical center,we find the equations of the radical axes by subtracting the circle equations.
Let $S_1: x^2+y^2-8x+40=0$
$S_2: x^2+y^2-5x+16=0$
$S_3: x^2+y^2-8x+16y+160=0$
Radical axis $S_1 - S_2 = 0$:
$(-8x+5x) + (40-16) = 0 \implies -3x + 24 = 0 \implies x = 8$.
Radical axis $S_1 - S_3 = 0$:
$(-8x+8x) - 16y + (40-160) = 0 \implies -16y - 120 = 0 \implies 16y = -120 \implies y = -\frac{120}{16} = -\frac{15}{2}$.
The radical center is the intersection of these axes,which is $\left(8, -\frac{15}{2}\right)$.

(C) Given circles are $S_1: x^2+y^2+4x+6y+7=0$ and $S_2: 4x^2+4y^2+8x+12y-9=0$.
To find the radical axis,we first normalize the equation of $S_2$ by dividing by $4$:
$S_2: x^2+y^2+2x+3y-\frac{9}{4}=0$.
The equation of the radical axis is given by $S_1 - S_2 = 0$.
$(x^2+y^2+4x+6y+7) - (x^2+y^2+2x+3y-\frac{9}{4}) = 0$.
$(4x-2x) + (6y-3y) + (7 + \frac{9}{4}) = 0$.
$2x + 3y + \frac{28+9}{4} = 0$.
$2x + 3y + \frac{37}{4} = 0$.
Multiplying by $4$,we get $8x + 12y + 37 = 0$.

(A) The radical axis of two circles $S_1=0$ and $S_2=0$ is given by $S_1 - S_2 = 0$.
$ (x^2+y^2-4x+6y-10) - (x^2+y^2+2x-6y+2) = 0 $
$ -6x + 12y - 12 = 0 $
$ x - 2y + 2 = 0 $
This is the equation of the radical axis.
To find the intersection with $S_1$,we substitute $x = 2y - 2$ into $S_1$:
$ (2y-2)^2 + y^2 - 4(2y-2) + 6y - 10 = 0 $
$ 4y^2 - 8y + 4 + y^2 - 8y + 8 + 6y - 10 = 0 $
$ 5y^2 - 10y + 2 = 0 $
The discriminant $D = b^2 - 4ac = (-10)^2 - 4(5)(2) = 100 - 40 = 60$.
Since $D > 0$,the quadratic equation has two real and distinct roots for $y$,which implies the radical axis cuts the circle $S_1$ at two real and distinct points.

(A) The radical axis of two circles $S_1=0$ and $S_2=0$ is given by $S_1-S_2=0$.
For $L_1$: $(x^2+y^2-4x-6y+5) - (x^2+y^2-2x-4y-1) = 0$.
Simplifying,we get $-2x-2y+6=0$,which is $x+y-3=0$.
For $L_2$: $(x^2+y^2+2x+2y-7) - (x^2+y^2+x+y+9) = 0$.
Simplifying,we get $x+y-16=0$.
The slopes of $L_1$ and $L_2$ are both $m = -1$.
Since the slopes are equal,$L_1$ is parallel to $L_2$.

(B) The radical axis of $S$ and $S'$ is given by $S - S' = 0$:
$(x^2 + y^2 + 2x + 17y + 4) - (x^2 + y^2 + 7x + 6y + 11) = 0$
$-5x + 11y - 7 = 0 \implies 5x - 11y + 7 = 0$ ...$(i)$
The radical axis of $S'$ and $S''$ is given by $S' - S'' = 0$:
$(x^2 + y^2 + 7x + 6y + 11) - (x^2 + y^2 - x + 22y + 3) = 0$
$8x - 16y + 8 = 0 \implies x - 2y + 1 = 0$ ...(ii)
Solving equations $(i)$ and (ii):
From (ii),$x = 2y - 1$. Substituting into $(i)$:
$5(2y - 1) - 11y + 7 = 0
10y - 5 - 11y + 7 = 0
-y + 2 = 0 \implies y = 2$.
Then $x = 2(2) - 1 = 3$.
The radical center is $(3, 2)$.
The length of the tangent from $(3, 2)$ to $S = 0$ is $\sqrt{S(3, 2)}$:
$\sqrt{3^2 + 2^2 + 2(3) + 17(2) + 4} = \sqrt{9 + 4 + 6 + 34 + 4} = \sqrt{57}$.

(B) Let the coordinates of point $P$ be $(h, k)$.
Given points are $A(-2, 1)$ and $B(3, 0)$.
The slope of $AP$ is $m_1 = \frac{k-1}{h+2}$.
The slope of $BP$ is $m_2 = \frac{k-0}{h-3} = \frac{k}{h-3}$.
Since $\angle APB = 90^{\circ}$,the product of the slopes must be $-1$,i.e.,$m_1 m_2 = -1$.
$\frac{k-1}{h+2} \times \frac{k}{h-3} = -1$.
$\frac{k^2-k}{h^2-h-6} = -1$.
$k^2-k = -(h^2-h-6)$.
$h^2+k^2-h-k-6 = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus of point $P$ is $x^2+y^2-x-y-6=0$.

(C) The normal at any point on a circle always passes through its center. The normal passes through $(4,3)$ and $(2,1)$.
The equation of this normal is $y-1 = \frac{3-1}{4-2}(x-2)$ $\Rightarrow y-1 = 1(x-2)$ $\Rightarrow y = x-1 \dots(1)$.
The center of the circle lies on this normal. We are also given that $2x-y-2=0$ is a diameter,so the center also lies on this line $\dots(2)$.
Solving $(1)$ and $(2)$ by substituting $y = x-1$ into $2x-(x-1)-2=0$,we get $x-1=0 \Rightarrow x=1$. Then $y=0$. Thus,the center is $(1,0)$.
The radius $r$ is the distance between the center $(1,0)$ and the point $(2,1)$ on the circle: $r = \sqrt{(2-1)^2 + (1-0)^2} = \sqrt{1^2+1^2} = \sqrt{2}$.
The equation of the circle with center $(1,0)$ and radius $\sqrt{2}$ is $(x-1)^2 + (y-0)^2 = (\sqrt{2})^2$ $\Rightarrow x^2-2x+1+y^2=2$ $\Rightarrow x^2+y^2-2x-1=0$.

(C) Let the coordinates of a point $P$ be $(h, k)$,which is the midpoint of the chord $AB$.
Now,$OP = \sqrt{(h-0)^2 + (k-0)^2} = \sqrt{h^2+k^2}$.
In $\triangle AOP$,the angle $\angle AOP = \frac{1}{2} \times \frac{2\pi}{3} = \frac{\pi}{3}$.
Using trigonometry in $\triangle AOP$,we have $\cos\left(\frac{\pi}{3}\right) = \frac{OP}{OA}$.
Since $OA$ is the radius of the circle,$OA = 3$.
$\Rightarrow \frac{1}{2} = \frac{\sqrt{h^2+k^2}}{3}$.
$\Rightarrow \sqrt{h^2+k^2} = \frac{3}{2}$.
Squaring both sides,we get $h^2+k^2 = \frac{9}{4}$.
Replacing $(h, k)$ with $(x, y)$,the required locus is $x^2+y^2 = \frac{9}{4}$.

(C) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$.
Since the circle passes through the origin $O(0, 0)$ and meets the axes at $A(a, 0)$ and $B(0, b)$, the segment $AB$ is the diameter of the circle.
The length of the diameter is $2 \times \text{radius} = 2 \times 6 = 12$.
Thus, $a^2 + b^2 = 12^2 = 144$.
Let $(h, k)$ be the centroid of $\triangle OAB$. The coordinates of the centroid are given by $h = \frac{0+a+0}{3} = \frac{a}{3}$ and $k = \frac{0+0+b}{3} = \frac{b}{3}$.
Therefore, $a = 3h$ and $b = 3k$.
Substituting these into the equation $a^2 + b^2 = 144$, we get $(3h)^2 + (3k)^2 = 144$.
$9h^2 + 9k^2 = 144$.
Dividing by $9$, we get $h^2 + k^2 = 16$.
Replacing $(h, k)$ with $(x, y)$, the locus of the centroid is $x^2 + y^2 = 16$.

(B) The condition for two points $(x_1, y_1)$ and $(x_2, y_2)$ to be conjugate with respect to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by the equation:
$x_1x_2 + y_1y_2 + g(x_1 + x_2) + f(y_1 + y_2) + c = 0$
Given the circle $x^2 + y^2 + 6x + 4y + 12 = 0$,we have $g = 3$,$f = 2$,and $c = 12$.
Substituting the points $(6, -k)$ and $(-3, 2)$ into the condition:
$(6)(-3) + (-k)(2) + 3(6 - 3) + 2(-k + 2) + 12 = 0$
$-18 - 2k + 3(3) + 2(-k + 2) + 12 = 0$
$-18 - 2k + 9 - 2k + 4 + 12 = 0$
$-4k + 7 = 0$
$4k = 7$
$k = \frac{7}{4}$

(A) The equation of the family of circles passing through the intersection of $S_1: x^2+y^2-1=0$ and $S_2: x^2+y^2-2x+y=0$ is given by $S_1 + kS_2 = 0$ for $k \neq -1$.
$(x^2+y^2-1) + k(x^2+y^2-2x+y) = 0$
$(1+k)x^2 + (1+k)y^2 - 2kx + ky - (1+k) = 0$
Dividing by $(1+k)$,we get $x^2 + y^2 - \frac{2k}{1+k}x + \frac{k}{1+k}y - 1 = 0$.
The center $(h, k')$ of this circle is given by $(-g, -f) = \left(\frac{k}{1+k}, \frac{-k}{2(1+k)}\right)$.
Let $x = \frac{k}{1+k}$ and $y = \frac{-k}{2(1+k)}$.
Then $2y = \frac{-k}{1+k}$.
Adding $x$ and $2y$,we get $x + 2y = \frac{k}{1+k} - \frac{k}{1+k} = 0$.
Thus,the locus is the line $x+2y=0$.

(C) We know that the sum of the squares of the direction cosines of a vector is $1$.
$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$
Given that the vector makes equal angles $\alpha$ with $x$ and $y$ axes,so $\alpha = \beta$. Also,it makes $90^{\circ}$ with the $z$-axis,so $\gamma = 90^{\circ}$.
Substituting these values into the identity:
$\cos^2 \alpha + \cos^2 \alpha + \cos^2 90^{\circ} = 1$
$2 \cos^2 \alpha + 0 = 1$
$\cos^2 \alpha = \frac{1}{2}$
$\cos \alpha = \pm \frac{1}{\sqrt{2}}$
Therefore,$\alpha = 45^{\circ}$ or $135^{\circ}$.

(A) Given that $D, E$,and $F$ are the mid-points of $AB, AC$,and $BC$ respectively in $\triangle ABC$.
Let the position vectors of vertices $A, B$,and $C$ be $\vec{a}, \vec{b}$,and $\vec{c}$ respectively.
Then,the position vectors of the mid-points are:
$\overrightarrow{D} = \frac{\vec{a} + \vec{b}}{2}$
$\overrightarrow{E} = \frac{\vec{a} + \vec{c}}{2}$
$\overrightarrow{F} = \frac{\vec{b} + \vec{c}}{2}$
Now,we calculate the vectors $\overrightarrow{BE}$ and $\overrightarrow{AF}$:
$\overrightarrow{BE} = \overrightarrow{E} - \overrightarrow{B} = \frac{\vec{a} + \vec{c}}{2} - \vec{b} = \frac{\vec{a} + \vec{c} - 2\vec{b}}{2}$
$\overrightarrow{AF} = \overrightarrow{F} - \overrightarrow{A} = \frac{\vec{b} + \vec{c}}{2} - \vec{a} = \frac{\vec{b} + \vec{c} - 2\vec{a}}{2}$
Adding these two vectors:
$\overrightarrow{BE} + \overrightarrow{AF} = \frac{\vec{a} + \vec{c} - 2\vec{b} + \vec{b} + \vec{c} - 2\vec{a}}{2}$
$= \frac{2\vec{c} - \vec{a} - \vec{b}}{2} = \vec{c} - \frac{\vec{a} + \vec{b}}{2}$
Since $\overrightarrow{DC} = \overrightarrow{C} - \overrightarrow{D} = \vec{c} - \frac{\vec{a} + \vec{b}}{2}$,we have:
$\overrightarrow{BE} + \overrightarrow{AF} = \overrightarrow{DC}$

(B) Given that,$\vec{a}=\frac{3}{2} \hat{k}$ and $\vec{b}=\frac{2 \hat{i}+2 \hat{j}-\hat{k}}{2} = \hat{i}+\hat{j}-\frac{1}{2} \hat{k}$.
First,calculate $\vec{a}+\vec{b} = \hat{i}+\hat{j}+(\frac{3}{2}-\frac{1}{2}) \hat{k} = \hat{i}+\hat{j}+\hat{k}$.
Next,calculate $\vec{a}-\vec{b} = -\hat{i}-\hat{j}+(\frac{3}{2}+\frac{1}{2}) \hat{k} = -\hat{i}-\hat{j}+2 \hat{k}$.
Let $\vec{u} = \vec{a}+\vec{b}$ and $\vec{v} = \vec{a}-\vec{b}$.
The dot product $\vec{u} \cdot \vec{v} = (1)(-1) + (1)(-1) + (1)(2) = -1 - 1 + 2 = 0$.
Since the dot product is $0$,the vectors are perpendicular.
Therefore,the angle $\theta$ between them is $90^{\circ}$.

(A) Let the diagonals of the parallelogram be $\overrightarrow{d_1} = \overrightarrow{a} + \overrightarrow{b}$ and $\overrightarrow{d_2} = \overrightarrow{b} + \overrightarrow{c}$.
$\overrightarrow{d_1} = (\hat{i} + \hat{j} + \hat{k}) + (\hat{i} + 3\hat{j} + 5\hat{k}) = 2\hat{i} + 4\hat{j} + 6\hat{k}$
$\overrightarrow{d_2} = (\hat{i} + 3\hat{j} + 5\hat{k}) + (7\hat{i} + 9\hat{j} + 11\hat{k}) = 8\hat{i} + 12\hat{j} + 16\hat{k}$
The area of a parallelogram with diagonals $\overrightarrow{d_1}$ and $\overrightarrow{d_2}$ is given by $\frac{1}{2} |\overrightarrow{d_1} \times \overrightarrow{d_2}|$.
$\overrightarrow{d_1} \times \overrightarrow{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 6 \\ 8 & 12 & 16 \end{vmatrix} = \hat{i}(64 - 72) - \hat{j}(32 - 48) + \hat{k}(24 - 32) = -8\hat{i} + 16\hat{j} - 8\hat{k}$
Magnitude $|\overrightarrow{d_1} \times \overrightarrow{d_2}| = \sqrt{(-8)^2 + 16^2 + (-8)^2} = \sqrt{64 + 256 + 64} = \sqrt{384} = \sqrt{64 \times 6} = 8\sqrt{6}$
Area $= \frac{1}{2} \times 8\sqrt{6} = 4\sqrt{6}$ sq units.

(D) vector $\vec{r} = a\hat{i} + b\hat{j} + c\hat{k}$ is equally inclined to the coordinate axes if its direction cosines are equal,i.e.,$\cos \alpha = \cos \beta = \cos \gamma$.
This implies that the direction ratios must be equal in magnitude,i.e.,$|a| = |b| = |c|$.
Checking the options:
For option $D$,$\vec{v} = 4\hat{i} + 4\hat{j} + 4\hat{k}$.
The direction ratios are $a=4, b=4, c=4$.
The magnitude is $|\vec{v}| = \sqrt{4^2 + 4^2 + 4^2} = \sqrt{48} = 4\sqrt{3}$.
The direction cosines are $\left(\frac{4}{4\sqrt{3}}, \frac{4}{4\sqrt{3}}, \frac{4}{4\sqrt{3}}\right) = \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$.
Since all direction cosines are equal,the vector $4\hat{i} + 4\hat{j} + 4\hat{k}$ is equally inclined with the coordinate axes.

(C) Given position vectors are $\vec{A} = \hat{i}+4 \hat{j}+3 \hat{k}$,$\vec{B} = \hat{i}+2 \hat{j}+3 \hat{k}$,and $\vec{C} = 3 \hat{i}+2 \hat{j}+\hat{k}$.
Since $D$ is the midpoint of $BC$,its position vector $\vec{D}$ is given by $\vec{D} = \frac{\vec{B} + \vec{C}}{2} = \frac{(\hat{i}+2 \hat{j}+3 \hat{k}) + (3 \hat{i}+2 \hat{j}+\hat{k})}{2} = \frac{4 \hat{i}+4 \hat{j}+4 \hat{k}}{2} = 2 \hat{i}+2 \hat{j}+2 \hat{k}$.
Since $E$ is the midpoint of $AC$,its position vector $\vec{E}$ is given by $\vec{E} = \frac{\vec{A} + \vec{C}}{2} = \frac{(\hat{i}+4 \hat{j}+3 \hat{k}) + (3 \hat{i}+2 \hat{j}+\hat{k})}{2} = \frac{4 \hat{i}+6 \hat{j}+4 \hat{k}}{2} = 2 \hat{i}+3 \hat{j}+2 \hat{k}$.
Now,$\overrightarrow{DE} = \vec{E} - \vec{D} = (2 \hat{i}+3 \hat{j}+2 \hat{k}) - (2 \hat{i}+2 \hat{j}+2 \hat{k}) = \hat{j}$.

(A) The position vectors of points $u$ and $v$ are given as $(2, 1)$ and $(3, -5)$ respectively.
The equation of the line passing through points $(x_1, y_1) = (2, 1)$ and $(x_2, y_2) = (3, -5)$ is given by $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.
Substituting the values: $y - 1 = \frac{-5 - 1}{3 - 2}(x - 2) \Rightarrow y - 1 = -6(x - 2) \Rightarrow y - 1 = -6x + 12 \Rightarrow 6x + y = 13$.
Now,we check which points satisfy the equation $6x + y = 13$:
For $P\left(\frac{5}{2}, -2\right)$: $6\left(\frac{5}{2}\right) + (-2) = 15 - 2 = 13$. (Satisfies)
For $Q\left(\frac{7}{3}, -1\right)$: $6\left(\frac{7}{3}\right) + (-1) = 14 - 1 = 13$. (Satisfies)
For $R\left(\frac{9}{4}, 0\right)$: $6\left(\frac{9}{4}\right) + 0 = \frac{27}{2} = 13.5 \neq 13$. (Does not satisfy)
Thus,only points $P$ and $Q$ lie on the line.

(B) We know that $|u-v|^2 = |u|^2 + |v|^2 - 2(\vec{u} \cdot \vec{v})$.
Also,$(||u|-|v||)^2 = |u|^2 + |v|^2 - 2|u||v|$.
For the equality $|u-v| = ||u|-|v||$ to hold,their squares must be equal:
$|u|^2 + |v|^2 - 2(\vec{u} \cdot \vec{v}) = |u|^2 + |v|^2 - 2|u||v|$.
This simplifies to $\vec{u} \cdot \vec{v} = |u||v|$.
Since $\vec{u} \cdot \vec{v} = |u||v| \cos \theta$,we have $|u||v| \cos \theta = |u||v|$.
This implies $\cos \theta = 1$,which means $\theta = 0$.
Therefore,$u$ and $v$ must have the same direction.

(A) Given that $\vec{OA}=\vec{a}$ and $\vec{OB}=\vec{b}$.
Since $B$ is the mid-point of $AC$,we have $\vec{OB} = \frac{\vec{OA} + \vec{OC}}{2}$.
Substituting the given values:
$\vec{b} = \frac{\vec{a} + \vec{OC}}{2}$
$2\vec{b} = \vec{a} + \vec{OC}$
$\vec{OC} = 2\vec{b} - \vec{a}$.

(D) By the triangle law of vector addition,we have:
$\vec{AB} + \vec{BC} = \vec{AC}$
Similarly,for the other side of the hexagon:
$\vec{FE} + \vec{ED} = \vec{FD}$
In a regular hexagon $ABCDEF$,the diagonal $\vec{AC}$ is parallel to the diagonal $\vec{FD}$.
Therefore,the vector $\vec{AB} + \vec{BC}$ is parallel to $\vec{FE} + \vec{ED}$.

(B) Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if their scalar triple product is zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = 0$.
The given vectors are $\vec{a} = 35 \hat{i}+14 \hat{j}-77 \hat{k}$,$\vec{b} = 2 \hat{i}+7 \hat{j}+5 \hat{k}$,and $\vec{c} = 5 \hat{i}+2 \hat{j}+\lambda \hat{k}$.
The condition for coplanarity is given by the determinant:
$\left|\begin{array}{ccc} 35 & 14 & -77 \\ 2 & 7 & 5 \\ 5 & 2 & \lambda \end{array}\right| = 0$
Expanding the determinant along the first row:
$35(7\lambda - 10) - 14(2\lambda - 25) - 77(4 - 35) = 0$
$245\lambda - 350 - 28\lambda + 350 - 77(-31) = 0$
$217\lambda + 2387 = 0$
Dividing by $217$:
$\lambda + 11 = 0$
$\lambda = -11$

(D) In a regular hexagon $ABCDEF$ with center $O$,the vector $\vec{AO}$ represents the directed line segment from the center $O$ to the vertex $A$.
By the properties of a regular hexagon,the vector $\vec{AO}$ is equal to the vector $\vec{ED}$ and $\vec{BC}$.
Looking at the options provided,$\vec{BC}$ is not explicitly listed,but we can analyze the vectors given.
Note that $\vec{AO} = \vec{ED} = \vec{BC}$.
Since $\vec{BC}$ is not an option,let us re-examine the geometry.
Actually,$\vec{AO} = \vec{ED}$ is correct.
Thus,the vector $\vec{AO}$ is equal to $\vec{ED}$.

(B) Let the vectors be $\vec{OA}, \vec{OB}, \vec{OC}$. We need to find the magnitude of $\vec{R} = \vec{OA} + \vec{OB} + \vec{OC}$.
Let $\vec{OB}$ be along the $y$-axis,so $\vec{OB} = R\hat{j}$.
Then $\vec{OA} = R(\cos 45^{\circ} \hat{i} + \sin 45^{\circ} \hat{j}) = R(\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j})$.
And $\vec{OC} = R(\cos 45^{\circ} \hat{i} - \sin 45^{\circ} \hat{j}) = R(\frac{1}{\sqrt{2}}\hat{i} - \frac{1}{\sqrt{2}}\hat{j})$.
Summing these vectors:
$\vec{R} = \vec{OA} + \vec{OB} + \vec{OC} = R(\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}) + R\hat{j} + R(\frac{1}{\sqrt{2}}\hat{i} - \frac{1}{\sqrt{2}}\hat{j})$
$\vec{R} = R(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})\hat{i} + R(\frac{1}{\sqrt{2}} + 1 - \frac{1}{\sqrt{2}})\hat{j}$
$\vec{R} = R(\frac{2}{\sqrt{2}})\hat{i} + R\hat{j} = \sqrt{2}R\hat{i} + R\hat{j}$.
The magnitude is $|\vec{R}| = \sqrt{(\sqrt{2}R)^2 + R^2} = \sqrt{2R^2 + R^2} = \sqrt{3R^2} = \sqrt{3}R$.
Wait,re-evaluating the geometry: The resultant of $\vec{OA}$ and $\vec{OC}$ is $2R \cos(45^{\circ}) \hat{j} = \sqrt{2}R \hat{j}$.
Adding $\vec{OB} = R \hat{j}$,the total resultant is $(\sqrt{2} + 1)R \hat{j}$.
Thus,the magnitude is $(\sqrt{2} + 1)R$.

(D) Let the position vectors of the four points be $A = \hat{i}+\hat{j}-\hat{k}$,$B = 2\hat{i}+3\hat{j}$,$C = 5\hat{j}-2\hat{k}$,and $D = -\hat{j}+\hat{k}$.
We calculate the vectors representing the sides:
$\vec{AB} = B - A = (2\hat{i}+3\hat{j}) - (\hat{i}+\hat{j}-\hat{k}) = \hat{i}+2\hat{j}+\hat{k}$.
$\vec{BC} = C - B = (5\hat{j}-2\hat{k}) - (2\hat{i}+3\hat{j}) = -2\hat{i}+2\hat{j}-2\hat{k}$.
$\vec{CD} = D - C = (-\hat{j}+\hat{k}) - (5\hat{j}-2\hat{k}) = -6\hat{j}+3\hat{k}$.
$\vec{DA} = A - D = (\hat{i}+\hat{j}-\hat{k}) - (-\hat{j}+\hat{k}) = \hat{i}+2\hat{j}-2\hat{k}$.
Since the vectors representing the sides are not parallel to each other (i.e.,$\vec{AB} \neq k\vec{CD}$ and $\vec{BC} \neq k\vec{DA}$),the figure does not satisfy the conditions for a parallelogram or a trapezium.
Thus,the figure formed by these four points is a general quadrilateral.

(B) Let $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}$.
The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by $|\vec{a} \times \vec{b}|$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ 3 & -2 & 1 \end{vmatrix}$
$= \hat{i}(1(1) - 2(-2)) - \hat{j}(1(1) - 2(3)) + \hat{k}(1(-2) - 1(3))$
$= \hat{i}(1 + 4) - \hat{j}(1 - 6) + \hat{k}(-2 - 3)$
$= 5\hat{i} + 5\hat{j} - 5\hat{k}$
Now,calculate the magnitude $|\vec{a} \times \vec{b}| = \sqrt{5^2 + 5^2 + (-5)^2}$
$= \sqrt{25 + 25 + 25} = \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$.
Thus,the area is $5\sqrt{3}$ square units.

(A) In parallelogram $ABCD$,we have $\vec{AB} = \vec{DC}$ and $\vec{AD} = \vec{BC}$.
Since $L$ is the mid-point of $BC$,we have $\vec{BL} = \frac{1}{2} \vec{BC}$.
Using the triangle law of vector addition in $\triangle ABL$,we have:
$\vec{AL} = \vec{AB} + \vec{BL}$
Substituting the known values:
$\vec{AL} = \vec{DC} + \frac{1}{2} \vec{BC}$
Since $\vec{BC} = \vec{AD}$ in a parallelogram,we get:
$\vec{AL} = \vec{DC} + \frac{1}{2} \vec{AD}$

(C) Two vectors $\vec{a} = a_1 \hat{i} + b_1 \hat{j} + c_1 \hat{k}$ and $\vec{b} = a_2 \hat{i} + b_2 \hat{j} + c_2 \hat{k}$ are collinear if their components are proportional,i.e.,$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = k$ for some constant $k$.
Given vectors are $\vec{a} = \hat{i} + 2 \hat{j} + m \hat{k}$ and $\vec{b} = \hat{i} + m \hat{j} + 2 \hat{k}$.
For these to be collinear,we must have $\frac{1}{1} = \frac{2}{m} = \frac{m}{2}$.
From $\frac{1}{1} = \frac{2}{m}$,we get $m = 2$.
From $\frac{1}{1} = \frac{m}{2}$,we get $m = 2$.
Since both conditions yield the same value $m = 2$,there is only $1$ such value of $m$ for which the vectors are collinear.

(A) Given that $ABCDEF$ is a regular hexagon with $\vec{AB} = \vec{a}$ and $\vec{BC} = \vec{b}$.
In a regular hexagon,the opposite sides are parallel and equal in magnitude. Thus,$\vec{ED} = \vec{AB} = \vec{a}$ and $\vec{FE} = \vec{BC} = \vec{b}$.
Also,the main diagonal $\vec{AD}$ is parallel to $\vec{BC}$ and its magnitude is twice that of $\vec{BC}$. Therefore,$\vec{AD} = 2\vec{BC} = 2\vec{b}$.
Using the triangle law of vector addition in $\triangle ABC$,we have:
$\vec{AC} = \vec{AB} + \vec{BC} = \vec{a} + \vec{b}$.
Now,in $\triangle ACD$,by the triangle law of vector addition:
$\vec{AC} + \vec{CD} = \vec{AD}$
Substituting the known values:
$(\vec{a} + \vec{b}) + \vec{CD} = 2\vec{b}$
$\vec{CD} = 2\vec{b} - (\vec{a} + \vec{b})$
$\vec{CD} = 2\vec{b} - \vec{a} - \vec{b}$
$\vec{CD} = \vec{b} - \vec{a}$

(D) Let the center of the regular hexagon $ABCDEF$ be the origin $O$.
We express each vector in terms of the position vectors of the vertices relative to the origin $O$.
$\vec{BE} + \vec{BC} + \vec{EF} + \vec{BA} + \vec{CF} + \vec{AF}$
$= (\vec{OE} - \vec{OB}) + (\vec{OC} - \vec{OB}) + (\vec{OF} - \vec{OE}) + (\vec{OA} - \vec{OB}) + (\vec{OF} - \vec{OC}) + (\vec{OF} - \vec{OA})$
Grouping the terms,we observe that $\vec{OE} - \vec{OE} = 0$,$\vec{OC} - \vec{OC} = 0$,and $\vec{OA} - \vec{OA} = 0$.
This simplifies to: $3\vec{OF} - 3\vec{OB}$
$= 3(\vec{OF} - \vec{OB})$
$= 3\vec{BF}$.

(C) In a regular hexagon $ABCDEF$,let $\vec{AB} = \vec{a}$ and $\vec{BC} = \vec{b}$.
Since it is a regular hexagon,$\vec{CD} = \vec{AF} = \vec{BC} - \vec{AB} = \vec{b} - \vec{a}$.
Also,$\vec{DE} = -\vec{AB} = -\vec{a}$.
Using the triangle law of vector addition in $\triangle CDE$,we have $\vec{CE} = \vec{CD} + \vec{DE}$.
Substituting the values,$\vec{CE} = (\vec{b} - \vec{a}) + (-\vec{a}) = \vec{b} - 2\vec{a}$.

(B) Given the vector equation: $PQ + QR = (2\lambda^2 - 5)RP$
By the triangle law of vector addition,we know that $PQ + QR = PR$.
Substituting this into the given equation,we get: $PR = (2\lambda^2 - 5)RP$
Since $PR = -RP$,we can rewrite the equation as: $-RP = (2\lambda^2 - 5)RP$
Dividing both sides by $RP$ (assuming $RP \neq 0$),we get: $2\lambda^2 - 5 = -1$
$2\lambda^2 = 4$
$\lambda^2 = 2$
$\lambda = \pm \sqrt{2}$

(C) Since $C$ is the mid-point of the line segment $AB$,the position vector of $C$ is given by $\vec{c} = \frac{\vec{a} + \vec{b}}{2}$.
In terms of vectors from point $P$,we have $\vec{PC} = \frac{\vec{PA} + \vec{PB}}{2}$.
Multiplying by $2$,we get $2\vec{PC} = \vec{PA} + \vec{PB}$.
Rearranging the terms,we can write $\vec{PA} - \vec{PC} = \vec{PC} - \vec{PB}$.
Thus,the correct relation is $\vec{PA} - \vec{PC} = \vec{PC} - \vec{PB}$.

(D) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$.
Consider the expression $|a - b|^2$:
$|a - b|^2 = |a|^2 + |b|^2 - 2(a \cdot b)$
Since $a \cdot b = |a||b| \cos \theta = 1 \times 1 \times \cos \theta = \cos \theta$,we have:
$|a - b|^2 = 1^2 + 1^2 - 2 \cos \theta$
$|a - b|^2 = 2 - 2 \cos \theta$
$|a - b|^2 = 2(1 - \cos \theta)$
Using the trigonometric identity $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$:
$|a - b|^2 = 2(2 \sin^2 \frac{\theta}{2})$
$|a - b|^2 = 4 \sin^2 \frac{\theta}{2}$
Taking the square root on both sides:
$|a - b| = 2 \sin \frac{\theta}{2}$
Therefore,$\sin \frac{\theta}{2} = \frac{1}{2} |a - b|$.

(D) Given that $a$ and $b$ are unit vectors,we have $|a| = 1$ and $|b| = 1$.
Also,$a+2b$ is a unit vector,so $|a+2b| = 1$.
Squaring both sides,we get $|a+2b|^2 = 1^2 = 1$.
Expanding the dot product,we have $|a|^2 + 4|b|^2 + 4(a \cdot b) = 1$.
Substituting the values,$1^2 + 4(1)^2 + 4|a||b| \cos \theta = 1$.
$1 + 4 + 4 \cos \theta = 1$.
$5 + 4 \cos \theta = 1$,which implies $4 \cos \theta = -4$,so $\cos \theta = -1$.
This means $\theta = \pi$.
Now,we evaluate the expression $\sin \theta + \cos^3 \theta + \tan^5 \theta$.
Substituting $\theta = \pi$: $\sin \pi + \cos^3 \pi + \tan^5 \pi = 0 + (-1)^3 + 0 = -1$.

(D) The vectors $u, v, w$ are coplanar if and only if their scalar triple product is zero,which is equivalent to the determinant of the matrix formed by their components being zero:
$\begin{vmatrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix} = 0$
Taking $a, b, c$ common from the first,second,and third columns respectively:
$abc \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} = 0$
The determinant is a standard Vandermonde determinant:
$abc(a-b)(b-c)(c-a) = 0$
This product is zero if $a=0$ or $b=0$ or $c=0$,or if $a=b$ or $b=c$ or $c=a$.
Thus,the vectors are coplanar if either one of $a, b, c$ is zero,or any two of $a, b, c$ are equal.

(D) Given that $a$,$b$,and $c$ are non-zero vectors such that no two of these are collinear.
Since $a + 2b$ is collinear with $c$,we have $a + 2b = m c$ for some non-zero scalar $m$ $(i)$.
Since $b + 3c$ is collinear with $a$,we have $b + 3c = n a$ for some non-zero scalar $n$ $(ii)$.
From $(ii)$,we have $b = n a - 3c$.
Substitute this into $(i)$:
$a + 2(n a - 3c) = m c$
$a + 2n a - 6c = m c$
$(1 + 2n) a = (m + 6) c$
Since $a$ and $c$ are non-collinear,the coefficients must be zero:
$1 + 2n = 0 \Rightarrow n = -\frac{1}{2}$
$m + 6 = 0 \Rightarrow m = -6$
Substituting $m = -6$ into equation $(i)$:
$a + 2b = -6c$
$a + 2b + 6c = 0$.

(B) Given vectors are $\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}$,$\vec{b}=\hat{i}+3 \hat{j}-\hat{k}$,and $\vec{c}=3 \hat{i}-\hat{j}-2 \hat{k}$.
First,we calculate the dot products:
$\vec{a} \cdot \vec{a} = (2)^2 + (1)^2 + (3)^2 = 4 + 1 + 9 = 14$
$\vec{b} \cdot \vec{b} = (1)^2 + (3)^2 + (-1)^2 = 1 + 9 + 1 = 11$
$\vec{c} \cdot \vec{c} = (3)^2 + (-1)^2 + (-2)^2 = 9 + 1 + 4 = 14$
$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} = (2)(1) + (1)(3) + (3)(-1) = 2 + 3 - 3 = 2$
$\vec{a} \cdot \vec{c} = \vec{c} \cdot \vec{a} = (2)(3) + (1)(-1) + (3)(-2) = 6 - 1 - 6 = -1$
$\vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{b} = (1)(3) + (3)(-1) + (-1)(-2) = 3 - 3 + 2 = 2$
Now,substitute these values into the determinant:
$\Delta = \left|\begin{array}{ccc} 14 & 2 & -1 \\ 2 & 11 & 2 \\ -1 & 2 & 14 \end{array}\right|$
Expanding along the first row:
$\Delta = 14(11 \times 14 - 2 \times 2) - 2(2 \times 14 - 2 \times (-1)) - 1(2 \times 2 - 11 \times (-1))$
$\Delta = 14(154 - 4) - 2(28 + 2) - 1(4 + 11)$
$\Delta = 14(150) - 2(30) - 1(15)$
$\Delta = 2100 - 60 - 15 = 2025$.

(C) We know that the dot product of two perpendicular vectors is zero.
Since $(\vec{a}+t \vec{b})$ and $(\vec{a}-t \vec{b})$ are perpendicular,their dot product is zero:
$(\vec{a}+t \vec{b}) \cdot (\vec{a}-t \vec{b}) = 0$
Expanding the dot product using the distributive property:
$|\vec{a}|^2 - t(\vec{a} \cdot \vec{b}) + t(\vec{b} \cdot \vec{a}) - t^2|\vec{b}|^2 = 0$
Since the dot product is commutative,$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$,so the middle terms cancel out:
$|\vec{a}|^2 - t^2|\vec{b}|^2 = 0$
Substituting the given values $|\vec{a}|=2$ and $|\vec{b}|=3$:
$2^2 - t^2(3^2) = 0$
$4 - 9t^2 = 0$
$9t^2 = 4$
$t^2 = \frac{4}{9}$
$t = \pm \frac{2}{3}$
Since $t$ is given as a positive scalar,we have $t = \frac{2}{3}$.

(B) Given,$|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|$.
Since $|\vec{a} \cdot \vec{b}| = |\vec{a}||\vec{b}| |\cos \theta|$ and $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| |\sin \theta|$,where $\theta$ is the angle between vectors $\vec{a}$ and $\vec{b}$,we have:
$|\vec{a}||\vec{b}| |\cos \theta| = |\vec{a}||\vec{b}| |\sin \theta|$
$|\cos \theta| = |\sin \theta|$
$|\tan \theta| = 1$
This implies $\theta = \frac{\pi}{4}$ or $\theta = \frac{3\pi}{4}$.
Given that $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} < 0$,we have $\cos \theta < 0$.
Since $\cos(\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}} < 0$,the angle is $\theta = \frac{3\pi}{4}$.
Note that $\sec^{-1}(-\sqrt{2}) = \frac{3\pi}{4}$ because $\sec(\frac{3\pi}{4}) = -\sqrt{2}$.

(C) Given,$\vec{a}=(\sin x) \hat{i}+(\sin y) \hat{j}$ and $\vec{b}=(\cos x) \hat{i}+(\cos y) \hat{j}$.
We calculate the cross product $\vec{a} \times \vec{b}$ as follows:
$\vec{a} \times \vec{b} = ((\sin x) \hat{i}+(\sin y) \hat{j}) \times ((\cos x) \hat{i}+(\cos y) \hat{j})$
$= (\sin x \cos y) (\hat{i} \times \hat{i}) + (\sin x \cos y) (\hat{i} \times \hat{j}) + (\sin y \cos x) (\hat{j} \times \hat{i}) + (\sin y \cos y) (\hat{j} \times \hat{j})$
Since $\hat{i} \times \hat{i} = 0$,$\hat{j} \times \hat{j} = 0$,$\hat{i} \times \hat{j} = \hat{k}$,and $\hat{j} \times \hat{i} = -\hat{k}$:
$\vec{a} \times \vec{b} = (\sin x \cos y) \hat{k} - (\sin y \cos x) \hat{k} = (\sin x \cos y - \cos x \sin y) \hat{k} = \sin(x-y) \hat{k}$.
Now,the magnitude is $|\vec{a} \times \vec{b}| = |\sin(x-y)|$.
Since the range of the sine function is $[-1, 1]$,the absolute value $|\sin(x-y)|$ must be in the interval $[0, 1]$.
Therefore,$|\vec{a} \times \vec{b}| \leq 1$.

(A) Given vectors are $\vec{u} = 2\hat{i} + 3\hat{j} + \hat{k}$,$\vec{v} = -3\hat{i} + 2\hat{j}$,and $\vec{w} = \hat{i} - \hat{j} + 4\hat{k}$.
Two vectors are perpendicular if their dot product is $0$.
First,calculate $\vec{u} \cdot \vec{v} = (2)(-3) + (3)(2) + (1)(0) = -6 + 6 + 0 = 0$. Since the dot product is $0$,$\vec{u} \perp \vec{v}$.
Next,calculate $\vec{u} \cdot \vec{w} = (2)(1) + (3)(-1) + (1)(4) = 2 - 3 + 4 = 3 \neq 0$. Thus,$\vec{u}$ is not perpendicular to $\vec{w}$.
Finally,calculate $\vec{v} \cdot \vec{w} = (-3)(1) + (2)(-1) + (0)(4) = -3 - 2 + 0 = -5 \neq 0$. Thus,$\vec{v}$ is not perpendicular to $\vec{w}$.
Therefore,$\vec{u}$ is perpendicular to $\vec{v}$ but not to $\vec{w}$.

(C) We know that for any two vectors $\vec{a}$ and $\vec{b}$,the magnitude of the cross product is $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$,where $\theta$ is the angle between them. Squaring both sides,we get $(\vec{a} \times \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta$.
Similarly,the dot product is $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$. Squaring both sides,we get $(\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta$.
Substituting these into the given expression:
$\frac{(\vec{a} \times \vec{b})^2 + (\vec{a} \cdot \vec{b})^2}{2|\vec{a}|^2 |\vec{b}|^2} = \frac{|\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta + |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta}{2|\vec{a}|^2 |\vec{b}|^2}$
$= \frac{|\vec{a}|^2 |\vec{b}|^2 (\sin^2 \theta + \cos^2 \theta)}{2|\vec{a}|^2 |\vec{b}|^2}$
Since $\sin^2 \theta + \cos^2 \theta = 1$,the expression simplifies to:
$= \frac{|\vec{a}|^2 |\vec{b}|^2 (1)}{2|\vec{a}|^2 |\vec{b}|^2} = \frac{1}{2}$.

(C) The magnitude of the cross product of two vectors $\vec{u}$ and $\vec{v}$ is given by $|\vec{u} \times \vec{v}| = |\vec{u}||\vec{v}| \sin \theta$,where $\theta$ is the angle between them.
Given $\theta = 45^{\circ}$,we have $|\vec{u} \times \vec{v}| = |\vec{u}||\vec{v}| \sin 45^{\circ} = |\vec{u}||\vec{v}| \left( \frac{1}{\sqrt{2}} \right)$.
The dot product of the two vectors is given by $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}| \cos \theta$.
For $\theta = 45^{\circ}$,$\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}| \cos 45^{\circ} = |\vec{u}||\vec{v}| \left( \frac{1}{\sqrt{2}} \right)$.
Since $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$,it follows that $|\vec{u} \times \vec{v}| = \vec{u} \cdot \vec{v}$.

(A) We are given $\vec{a} = 3 \hat{i} - \hat{j}$ and $\vec{b} = 2 \hat{i} + \hat{j} - 3 \hat{k}$.
Since $\overrightarrow{b_1}$ is parallel to $\vec{a}$,we can write $\overrightarrow{b_1} = \lambda \vec{a} = \lambda(3 \hat{i} - \hat{j}) = 3\lambda \hat{i} - \lambda \hat{j}$.
We know $\vec{b} = \overrightarrow{b_1} + \overrightarrow{b_2}$,so $\overrightarrow{b_2} = \vec{b} - \overrightarrow{b_1} = (2 - 3\lambda) \hat{i} + (1 + \lambda) \hat{j} - 3 \hat{k}$.
Since $\overrightarrow{b_2}$ is perpendicular to $\vec{a}$,their dot product is zero: $\overrightarrow{b_2} \cdot \vec{a} = 0$.
$(2 - 3\lambda)(3) + (1 + \lambda)(-1) + (-3)(0) = 0$.
$6 - 9\lambda - 1 - \lambda = 0$.
$5 - 10\lambda = 0 \Rightarrow \lambda = \frac{1}{2}$.
Substituting $\lambda = \frac{1}{2}$ into the expression for $\overrightarrow{b_2}$:
$\overrightarrow{b_2} = (2 - 3(\frac{1}{2})) \hat{i} + (1 + \frac{1}{2}) \hat{j} - 3 \hat{k}$.
$\overrightarrow{b_2} = (2 - \frac{3}{2}) \hat{i} + (\frac{3}{2}) \hat{j} - 3 \hat{k} = \frac{1}{2} \hat{i} + \frac{3}{2} \hat{j} - 3 \hat{k}$.

(A) Given,$a, b, c, d$ are vectors such that $|d|=1$.
The given equations are:
$a+b+c=s d$ $(1)$
$b+c+d=a$ $(2)$
From equation $(2)$,we can write $b+c = a-d$.
Substituting this into equation $(1)$:
$a + (a-d) = s d$
$2a - d = s d$
$2a = (s+1) d$
$a = \frac{s+1}{2} d$
Given that $a \cdot d = 4$. Substituting the value of $a$:
$\left(\frac{s+1}{2} d\right) \cdot d = 4$
Since $|d|=1$,we have $d \cdot d = |d|^2 = 1^2 = 1$.
$\frac{s+1}{2} (1) = 4$
$s+1 = 8$
$s = 7$.

(B) Given,$A = 2 \hat{i} + \log_2 x \hat{j} + s \hat{k}$ and $B = \log_2 x \hat{i} + s \hat{j} + \log_2 x \hat{k}$.
Let the angle between $A$ and $B$ be $\theta$. Then,$\cos \theta = \frac{A \cdot B}{|A| |B|}$.
$A \cdot B = (2)(\log_2 x) + (\log_2 x)(s) + (s)(\log_2 x) = 2 \log_2 x + 2s \log_2 x = 2 \log_2 x (1 + s)$.
For $\theta$ to be an acute angle,we must have $\cos \theta > 0$.
Since $|A| > 0$ and $|B| > 0$,the condition $\cos \theta > 0$ implies $A \cdot B > 0$.
Therefore,$2 \log_2 x (1 + s) > 0$.
Given $\log_2 x > 0$,we can divide by $2 \log_2 x$ without changing the inequality sign.
Thus,$1 + s > 0$,which implies $s > -1$.

(D) Given $A = \hat{i} + 2 \hat{j}$.
Let $B = x \hat{i} + y \hat{j}$.
We are given the equations $(A + B) \cdot B = 15$ and $A \cdot B = 6$.
Expanding the first equation: $A \cdot B + B \cdot B = 15$.
Since $B \cdot B = |B|^2$,we have $A \cdot B + |B|^2 = 15$.
Substituting the value of $A \cdot B = 6$:
$6 + |B|^2 = 15$.
$|B|^2 = 15 - 6 = 9$.
Therefore,$|B| = \sqrt{9} = 3$.

(A) Given,$|a| = |b| = |c| = 1$ and $a \cdot b = 0 = a \cdot c$. The angle between $b$ and $c$ is $\frac{\pi}{3}$.
We need to find $|a \times b - a \times c|$.
Using the distributive property of the cross product,we have $|a \times b - a \times c| = |a \times (b - c)|$.
Since $a \cdot b = 0$ and $a \cdot c = 0$,we have $a \cdot (b - c) = 0$,which implies that $a$ is perpendicular to $(b - c)$.
Thus,$|a \times (b - c)| = |a| |b - c| \sin \frac{\pi}{2} = |a| |b - c| (1) = |b - c|$.
Now,calculate $|b - c|^2 = |b|^2 + |c|^2 - 2(b \cdot c) = 1 + 1 - 2(|b| |c| \cos \frac{\pi}{3}) = 2 - 2(1 \cdot 1 \cdot \frac{1}{2}) = 2 - 1 = 1$.
Therefore,$|b - c| = 1$.
Hence,$|a \times b - a \times c| = 1$.

(B) Given that $a, b, c$ are three vectors such that $|a|=3, |b|=4, |c|=5$ and $a+b+c=0$.
We have $a+b+c=0$,which implies $a+b=-c$.
Squaring both sides,we get $(a+b)^2 = (-c)^2$.
Expanding the dot product,we have $|a|^2 + |b|^2 + 2(a \cdot b) = |c|^2$.
Substituting the given magnitudes: $3^2 + 4^2 + 2(a \cdot b) = 5^2$.
$9 + 16 + 2(a \cdot b) = 25$.
$25 + 2(a \cdot b) = 25$.
$2(a \cdot b) = 0$.
Therefore,$a \cdot b = 0$.

(D) Given vectors are $a = x^2 \hat{i} + x \hat{j} + 3 \hat{k}$ and $b = x \hat{i} - 4 \hat{j} + 2 \hat{k}$.
We are given the condition $a \cdot b > 6$.
The dot product of two vectors $a = a_1 \hat{i} + b_1 \hat{j} + c_1 \hat{k}$ and $b = a_2 \hat{i} + b_2 \hat{j} + c_2 \hat{k}$ is defined as $a \cdot b = a_1 a_2 + b_1 b_2 + c_1 c_2$.
Substituting the given vectors:
$(x^2 \hat{i} + x \hat{j} + 3 \hat{k}) \cdot (x \hat{i} - 4 \hat{j} + 2 \hat{k}) > 6$
$x^2(x) + x(-4) + 3(2) > 6$
$x^3 - 4x + 6 > 6$
$x^3 - 4x > 0$
$x(x^2 - 4) > 0$
$x(x - 2)(x + 2) > 0$
Using the wavy curve method,the critical points are $x = -2, 0, 2$.
Testing the intervals:
For $x > 2$,$x(x-2)(x+2) > 0$ (Positive).
For $0 < x < 2$,$x(x-2)(x+2) < 0$ (Negative).
For $-2 < x < 0$,$x(x-2)(x+2) > 0$ (Positive).
For $x < -2$,$x(x-2)(x+2) < 0$ (Negative).
Thus,the solution is $x \in (-2, 0) \cup (2, \infty)$.

(B) Given,$A(4,3,5)$,$B(0,6,0)$,$C(-8,1,4)$ and $D$ are the vertices of a parallelogram.
Let $D$ be the point $(x, y, z)$.
Since the diagonals of a parallelogram bisect each other,the mid-point of $AC$ is equal to the mid-point of $BD$.
$\left(\frac{4+(-8)}{2}, \frac{3+1}{2}, \frac{5+4}{2}\right) = \left(\frac{x+0}{2}, \frac{y+6}{2}, \frac{z+0}{2}\right)$
$\left(-2, 2, \frac{9}{2}\right) = \left(\frac{x}{2}, \frac{y+6}{2}, \frac{z}{2}\right)$
Equating the coordinates,we get $x = -4$,$y = -2$,$z = 9$.
Thus,$D$ is $(-4, -2, 9)$.
Now,the vector $\vec{AC} = (-8-4)\hat{i} + (1-3)\hat{j} + (4-5)\hat{k} = -12\hat{i} - 2\hat{j} - \hat{k}$.
The vector $\vec{BD} = (-4-0)\hat{i} + (-2-6)\hat{j} + (9-0)\hat{k} = -4\hat{i} - 8\hat{j} + 9\hat{k}$.
Let $\theta$ be the angle between $\vec{AC}$ and $\vec{BD}$.
$\cos \theta = \frac{|\vec{AC} \cdot \vec{BD}|}{|\vec{AC}| |\vec{BD}|} = \frac{|(-12)(-4) + (-2)(-8) + (-1)(9)|}{\sqrt{(-12)^2 + (-2)^2 + (-1)^2} \sqrt{(-4)^2 + (-8)^2 + 9^2}}$
$\cos \theta = \frac{|48 + 16 - 9|}{\sqrt{144 + 4 + 1} \sqrt{16 + 64 + 81}} = \frac{55}{\sqrt{149} \sqrt{161}}$
Therefore,$\theta = \cos ^{-1}\left(\frac{55}{\sqrt{149} \sqrt{161}}\right)$.

(D) Given,$|a| = 7$ and $|b| = 8$.
The dot product of two vectors is given by $a \cdot b = |a||b| \cos \theta$,where $\theta$ is the angle between the vectors.
Taking the magnitude on both sides,we get $|a \cdot b| = |a||b| |\cos \theta|$.
Substituting the given values,$|a \cdot b| = 7 \times 8 |\cos \theta| = 56 |\cos \theta|$.
The maximum value of $|\cos \theta|$ is $1$,which occurs when $\theta = 0$ or $\theta = \pi$.
Therefore,the maximum value of $|a \cdot b|$ is $56 \times 1 = 56$.

(B) Given vectors are $a = \hat{i} + \hat{j}$ and $b = \hat{j} + \hat{k}$.
First,we find the cross product $a \times b$:
$a \times b = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} = \hat{i}(1-0) - \hat{j}(1-0) + \hat{k}(1-0) = \hat{i} - \hat{j} + \hat{k}$.
The magnitude of the cross product is $|a \times b| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.
The unit vectors perpendicular to both $a$ and $b$ are given by $\pm \frac{a \times b}{|a \times b|} = \pm \frac{1}{\sqrt{3}}(\hat{i} - \hat{j} + \hat{k})$.
Thus,there are exactly $2$ such unit vectors.

(A) Given vectors are $a=\hat{i}-\hat{j}+2 \hat{k}$,$b=2 \hat{i}+4 \hat{j}+\hat{k}$,and $c=\lambda \hat{i}+\hat{j}+\mu \hat{k}$.
Since the vectors are mutually orthogonal,their dot products are zero:
$a \cdot c = 0 \implies (1)(\lambda) + (-1)(1) + (2)(\mu) = 0 \implies \lambda + 2\mu = 1$ ...$(i)$
$b \cdot c = 0 \implies (2)(\lambda) + (4)(1) + (1)(\mu) = 0 \implies 2\lambda + \mu = -4$ ...(ii)
From equation (ii),we get $\mu = -4 - 2\lambda$.
Substituting this into equation $(i)$:
$\lambda + 2(-4 - 2\lambda) = 1$
$\lambda - 8 - 4\lambda = 1$
$-3\lambda = 9 \implies \lambda = -3$.
Now,substitute $\lambda = -3$ into $\mu = -4 - 2\lambda$:
$\mu = -4 - 2(-3) = -4 + 6 = 2$.
Thus,$(\lambda, \mu) = (-3, 2)$.

(B) Given vectors are $a = t^2 \hat{i} + e^t \hat{j} + \hat{k}$ and $b = 2 \hat{i} + t^2 \hat{j} + \log t \hat{k}$.
We know that the dot product $f(t) = a \cdot b$ is given by the sum of the products of corresponding components:
$f(t) = (t^2)(2) + (e^t)(t^2) + (1)(\log t) = 2t^2 + t^2 e^t + \log t$.
Now,differentiate $f(t)$ with respect to $t$ using the product rule and the derivative of $\log t$:
$f^{\prime}(t) = \frac{d}{dt}(2t^2) + \frac{d}{dt}(t^2 e^t) + \frac{d}{dt}(\log t)$
$f^{\prime}(t) = 4t + (2t e^t + t^2 e^t) + \frac{1}{t}$.
To find $f^{\prime}(1)$,substitute $t = 1$ into the derivative expression:
$f^{\prime}(1) = 4(1) + (2(1) e^1 + (1)^2 e^1) + \frac{1}{1}$
$f^{\prime}(1) = 4 + 2e + e + 1$
$f^{\prime}(1) = 5 + 3e$.

(B) Given that $a \times b = c$ and $b \times c = a$.
Since $a \times b = c$,the vector $c$ is perpendicular to both $a$ and $b$.
Since $b \times c = a$,the vector $a$ is perpendicular to both $b$ and $c$.
This implies that $a, b, c$ form an orthogonal set of vectors.
Using the property of the cross product,we know that $a \times c = a \times (a \times b)$.
Using the vector triple product identity,$a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$.
Here,$a \times c = -(c \times a)$.
Since $b \times c = a$,we have $a \times c = a \times (a \times b) = (a \cdot b)a - (a \cdot a)b$.
Since $a, b, c$ are mutually perpendicular,$a \cdot b = 0$.
Thus,$a \times c = -|a|^2 b$.
This shows that $a \times c$ is a scalar multiple of $b$,which means $a \times c$ is parallel to $b$.

(B) Given,$|a| = |b| = 2$,$a \cdot b = 2$ and $a + b + c = 0$.
We know that $a + b + c = 0 \implies a + b = -c$.
Squaring both sides,we get $|a + b|^2 = |-c|^2$.
$|a|^2 + |b|^2 + 2(a \cdot b) = |c|^2$.
Substituting the given values:
$2^2 + 2^2 + 2(2) = |c|^2$.
$4 + 4 + 4 = |c|^2$.
$|c|^2 = 12$.
$|c| = \sqrt{12} = 2 \sqrt{3}$.

(C) Given that $u$ and $v$ are non-zero vectors in $\mathbb{R}^3$.
We know that the magnitude of the cross product is $|u \times v| = |u||v| \sin \theta$ and the dot product is $u \cdot v = |u||v| \cos \theta$,where $\theta$ is the angle between the vectors $u$ and $v$.
Substituting these into the expression $|u \times v|^2 + |u \cdot v|^2$:
$|u \times v|^2 + |u \cdot v|^2 = (|u||v| \sin \theta)^2 + (|u||v| \cos \theta)^2$
$= |u|^2|v|^2 \sin^2 \theta + |u|^2|v|^2 \cos^2 \theta$
$= |u|^2|v|^2 (\sin^2 \theta + \cos^2 \theta)$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we get:
$= |u|^2|v|^2(1) = |u|^2|v|^2$.

(B) Since $u$ and $v$ are non-collinear vectors in $\mathbb{R}^2$,they are linearly independent and form a basis for $\mathbb{R}^2$. Thus,any vector in $\mathbb{R}^2$ can be expressed as a linear combination of $u$ and $v$. Statement $(i)$ is true.
By definition,the orthogonal projection of $u$ on $v$ is given by $w = \left( \frac{u \cdot v}{|v|^2} \right) v$. This is a scalar multiple of $v$. Since $w$ is a multiple of $v$,it can be written as $w = 0u + \left( \frac{u \cdot v}{|v|^2} \right) v$. For $w$ to be written as $au + bv$ with $a \neq 0$ and $b \neq 0$,$w$ would need to have a non-zero component in the direction of $u$. However,$w$ is orthogonal to $u - w$,and $w$ is parallel to $v$. Since $u$ and $v$ are non-collinear,$w$ cannot be expressed as $au + bv$ with $a \neq 0$. Thus,statement (ii) is false.

(D) Let the $YZ$-plane divide the line segment joining the points $A(2, 4, 5)$ and $B(3, 5, -4)$ in the ratio $k:1$.
Using the section formula,the coordinates of the point of division are given by $\left( \frac{3k+2}{k+1}, \frac{5k+4}{k+1}, \frac{-4k+5}{k+1} \right)$.
Since the point lies on the $YZ$-plane,its $x$-coordinate must be zero.
Therefore,$\frac{3k+2}{k+1} = 0$,which implies $3k + 2 = 0$,or $k = -\frac{2}{3}$.
Since the ratio $k:1$ is $-\frac{2}{3}:1$,which is equivalent to $-2:3$,the negative sign indicates that the division is external.
Thus,the $YZ$-plane divides the line segment in the ratio $2:3$ externally.