MHT CET 2024 Mathematics Question Paper with Answer and Solution

(C) The normal vectors to the given planes are $\vec{n}_1 = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{n}_2 = \hat{i} + 3\hat{j} + 2\hat{k}$.
Since the line $L$ is the intersection of these two planes,it is perpendicular to both normal vectors.
Therefore,the direction vector $\vec{v}$ of the line $L$ is given by the cross product $\vec{n}_1 \times \vec{n}_2$:
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 3 & 2 \end{vmatrix} = \hat{i}(6-3) - \hat{j}(4-1) + \hat{k}(6-3) = 3\hat{i} - 3\hat{j} + 3\hat{k}$.
We can simplify the direction vector to $\vec{u} = \hat{i} - \hat{j} + \hat{k}$.
The angle $\alpha$ that the line makes with the positive $X$-axis (represented by the unit vector $\hat{i}$) is given by $\cos \alpha = \frac{\vec{u} \cdot \hat{i}}{|\vec{u}| |\hat{i}|}$.
Calculating the dot product: $\vec{u} \cdot \hat{i} = (1)(1) + (-1)(0) + (1)(0) = 1$.
Calculating the magnitude: $|\vec{u}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.
Thus,$\cos \alpha = \frac{1}{\sqrt{3} \times 1} = \frac{1}{\sqrt{3}}$.

(B) $1$. Let the equation of the plane passing through $(1, 2, 1)$ be $a(x - 1) + b(y - 2) + c(z - 1) = 0$.
$2$. The plane is perpendicular to the planes $2x - 2y + z = 0$ and $x - y + 2z = 4$. The normal vectors are $\vec{n_1} = (2, -2, 1)$ and $\vec{n_2} = (1, -1, 2)$.
$3$. The normal vector $\vec{n}$ of the required plane is $\vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & -1 & 2 \end{vmatrix} = \hat{i}(-4 + 1) - \hat{j}(4 - 1) + \hat{k}(-2 + 2) = -3\hat{i} - 3\hat{j} + 0\hat{k}$.
$4$. We can take the normal vector as $(1, 1, 0)$. The equation of the plane is $1(x - 1) + 1(y - 2) + 0(z - 1) = 0$,which simplifies to $x + y - 3 = 0$.
$5$. The distance $d$ from the point $(2, 3, 4)$ to the plane $x + y - 3 = 0$ is $d = \frac{|2 + 3 - 3|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$ units.

(C) The equation of the $yz$-plane is $x = 0$. Since the required plane is perpendicular to the $yz$-plane,its normal vector must be perpendicular to the normal vector of the $yz$-plane (which is $\hat{i} = (1, 0, 0)$).
Let the points be $A(-1, 2, -2)$ and $B(-1, 3, 2)$.
The vector $\vec{AB} = (-1 - (-1))\hat{i} + (3 - 2)\hat{j} + (2 - (-2))\hat{k} = 0\hat{i} + 1\hat{j} + 4\hat{k}$.
Since the plane is perpendicular to the $yz$-plane,its normal vector $\vec{n}$ is perpendicular to $\hat{i} = (1, 0, 0)$ and $\vec{AB} = (0, 1, 4)$.
Thus,$\vec{n} = \hat{i} \times \vec{AB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 0 \\ 0 & 1 & 4 \end{vmatrix} = \hat{i}(0) - \hat{j}(4) + \hat{k}(1) = (0, -4, 1)$.
The equation of the plane passing through $(-1, 2, -2)$ with normal vector $(0, -4, 1)$ is:
$0(x + 1) - 4(y - 2) + 1(z + 2) = 0$
$-4y + 8 + z + 2 = 0$
$-4y + z + 10 = 0$
$4y - z = 10$.

(A) The equation of a plane passing through a point $\vec{a}$ and parallel to two lines with direction vectors $\vec{b}$ and $\vec{c}$ is given by $(\vec{r} - \vec{a}) \cdot (\vec{b} \times \vec{c}) = 0$,or $\vec{r} \cdot (\vec{b} \times \vec{c}) = \vec{a} \cdot (\vec{b} \times \vec{c})$.
Here,the point is $\vec{a} = 2\hat{i} - \hat{j} - 3\hat{k}$ and the direction vectors of the lines are $\vec{b} = 3\hat{i} + 2\hat{j} - 4\hat{k}$ and $\vec{c} = 2\hat{i} - 3\hat{j} + 2\hat{k}$.
First,calculate the normal vector $\vec{n} = \vec{b} \times \vec{c}$:
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -4 \\ 2 & -3 & 2 \end{vmatrix} = \hat{i}(4 - 12) - \hat{j}(6 - (-8)) + \hat{k}(-9 - 4) = -8\hat{i} - 14\hat{j} - 13\hat{k}$.
Now,calculate $\vec{a} \cdot \vec{n}$:
$\vec{a} \cdot \vec{n} = (2)(-8) + (-1)(-14) + (-3)(-13) = -16 + 14 + 39 = 37$.
The equation of the plane is $\vec{r} \cdot (-8\hat{i} - 14\hat{j} - 13\hat{k}) = 37$,which can be written as $-8x - 14y - 13z = 37$,or $8x + 14y + 13z + 37 = 0$.

(D) Let the point be $\vec{a} = \hat{i} + 2 \hat{j} - \hat{k}$ and the parallel vectors be $\vec{b} = 2 \hat{i} + \hat{j} - \hat{k}$ and $\vec{c} = \hat{i} - \hat{j} + 3 \hat{k}$.
The normal vector $\vec{n}$ to the plane is given by $\vec{n} = \vec{b} \times \vec{c}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & -1 & 3 \end{vmatrix} = \hat{i}(3 - 1) - \hat{j}(6 + 1) + \hat{k}(-2 - 1) = 2 \hat{i} - 7 \hat{j} - 3 \hat{k}$.
The vector equation of the plane is $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.
$\vec{a} \cdot \vec{n} = (\hat{i} + 2 \hat{j} - \hat{k}) \cdot (2 \hat{i} - 7 \hat{j} - 3 \hat{k}) = (1)(2) + (2)(-7) + (-1)(-3) = 2 - 14 + 3 = -9$.
Thus,the equation is $\vec{r} \cdot (2 \hat{i} - 7 \hat{j} - 3 \hat{k}) = -9$.

(C) The perpendicular distance $d$ of a point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by the formula:
$d = \left| \frac{Ax_1 + By_1 + Cz_1 + D}{\sqrt{A^2 + B^2 + C^2}} \right|$
Here,the point is the origin $(0, 0, 0)$ and the plane is $2x + y - 2z - 18 = 0$.
Substituting the values:
$d = \left| \frac{2(0) + 1(0) - 2(0) - 18}{\sqrt{2^2 + 1^2 + (-2)^2}} \right|$
$d = \left| \frac{-18}{\sqrt{4 + 1 + 4}} \right|$
$d = \left| \frac{-18}{\sqrt{9}} \right|$
$d = \left| \frac{-18}{3} \right|$
$d = |-6| = 6 \text{ units}$

(B) The equation of the line passing through $(1, -5, 9)$ and parallel to the line $x = y = z$ (direction ratios $1, 1, 1$) is given by $\frac{x - 1}{1} = \frac{y + 5}{1} = \frac{z - 9}{1} = \lambda$.
Any point on this line is of the form $(\lambda + 1, \lambda - 5, \lambda + 9)$.
Since this point lies on the plane $x - y + z = 5$,we substitute these coordinates into the plane equation:
$(\lambda + 1) - (\lambda - 5) + (\lambda + 9) = 5$.
$\lambda + 1 - \lambda + 5 + \lambda + 9 = 5$.
$\lambda + 15 = 5$,which gives $\lambda = -10$.
The point of intersection is $(-10 + 1, -10 - 5, -10 + 9) = (-9, -15, -1)$.
The required distance is the distance between $(1, -5, 9)$ and $(-9, -15, -1)$:
$d = \sqrt{(-9 - 1)^2 + (-15 - (-5))^2 + (-1 - 9)^2}$.
$d = \sqrt{(-10)^2 + (-10)^2 + (-10)^2} = \sqrt{100 + 100 + 100} = \sqrt{300} = 10 \sqrt{3}$ units.

(C) Let the equation of the variable plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Since it passes through the fixed point $(3, 2, 1)$,we have $\frac{3}{a} + \frac{2}{b} + \frac{1}{c} = 1$ (Equation $i$).
The coordinates of points $A, B,$ and $C$ are $(a, 0, 0), (0, b, 0),$ and $(0, 0, c)$ respectively.
The plane passing through $A(a, 0, 0)$ parallel to the $YZ$-plane is $x = a$.
The plane passing through $B(0, b, 0)$ parallel to the $ZX$-plane is $y = b$.
The plane passing through $C(0, 0, c)$ parallel to the $XY$-plane is $z = c$.
The point of intersection of these three planes is $(x, y, z) = (a, b, c)$.
Substituting $a = x, b = y,$ and $c = z$ into Equation $i$,we get the locus: $\frac{3}{x} + \frac{2}{y} + \frac{1}{z} = 1$.

(A) Let $A = (5, -1, 4)$ and $B = (4, -1, 3)$.
The vector $\vec{AB} = (4-5)\hat{i} + (-1-(-1))\hat{j} + (3-4)\hat{k} = -\hat{i} - \hat{k}$.
The magnitude of the vector is $|\vec{AB}| = \sqrt{(-1)^2 + 0^2 + (-1)^2} = \sqrt{2}$.
The normal vector to the plane $x+y+z=7$ is $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.
Let $\theta$ be the angle between the line segment $AB$ and the plane. The angle $\phi$ between the line segment and the normal to the plane is given by $\cos \phi = \frac{|\vec{AB} \cdot \vec{n}|}{|\vec{AB}| |\vec{n}|}$.
$\cos \phi = \frac{|(-1)(1) + (0)(1) + (-1)(1)|}{\sqrt{2} \sqrt{1^2+1^2+1^2}} = \frac{|-2|}{\sqrt{2} \sqrt{3}} = \frac{2}{\sqrt{6}} = \sqrt{\frac{2}{3}}$.
Since $\phi$ is the angle with the normal,the angle $\theta$ with the plane is $90^\circ - \phi$,so $\sin \theta = \cos \phi = \sqrt{\frac{2}{3}}$.
Then $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}}$.
The length of the projection of the line segment on the plane is $|\vec{AB}| \cos \theta = \sqrt{2} \times \sqrt{\frac{1}{3}} = \sqrt{\frac{2}{3}}$ units.

(D) The equation of the plane containing the two given lines is given by the determinant form:
$\begin{vmatrix} x-1 & y-2 & z-3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = 0$
Expanding along the first row:
$(x-1)(15-16) - (y-2)(10-12) + (z-3)(8-9) = 0$
$-(x-1) + 2(y-2) - (z-3) = 0$
$-x + 1 + 2y - 4 - z + 3 = 0$
$-x + 2y - z = 0 \implies x - 2y + z = 0$ ... $(i)$
Given the plane equation $Ax - 2y + z = d$ ... $(ii)$
Since the planes are parallel,the coefficients of $x, y, z$ must be proportional. Thus,$A = 1$.
The distance $D$ between two parallel planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$ is given by $D = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}$.
Here,$D = \sqrt{6}$,$a=1, b=-2, c=1$,$d_1 = 0$,and $d_2 = d$.
$\sqrt{6} = \frac{|0 - d|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{|d|}{\sqrt{6}}$
$|d| = \sqrt{6} \times \sqrt{6} = 6$.

(B) The equation of any plane passing through the intersection of the planes $P_1: x+y+z-1=0$ and $P_2: 2x+3y-z+4=0$ is given by $P_1 + \lambda P_2 = 0$.
$(x+y+z-1) + \lambda(2x+3y-z+4) = 0$
$(1+2\lambda)x + (1+3\lambda)y + (1-\lambda)z + (4\lambda-1) = 0$
Since the plane is parallel to the $Y$-axis,its normal vector must be perpendicular to the $Y$-axis (which has direction vector $\vec{j} = (0, 1, 0)$).
Therefore,the coefficient of $y$ must be zero:
$1+3\lambda = 0 \Rightarrow \lambda = -\frac{1}{3}$
Substituting $\lambda = -\frac{1}{3}$ into the equation:
$(1 + 2(-\frac{1}{3}))x + (1 + 3(-\frac{1}{3}))y + (1 - (-\frac{1}{3}))z + (4(-\frac{1}{3}) - 1) = 0$
$(1 - \frac{2}{3})x + 0y + (1 + \frac{1}{3})z + (-\frac{4}{3} - 1) = 0$
$\frac{1}{3}x + \frac{4}{3}z - \frac{7}{3} = 0$
Multiplying by $3$,we get $x+4z-7=0$.

(D) The intercept form of the equation of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Given $a = 1$ and $b = 1$,the equation becomes $x + y + \frac{z}{c} = 1$.
Since the plane passes through $(-1, 1, 2)$,we substitute these coordinates: $-1 + 1 + \frac{2}{c} = 1$,which gives $\frac{2}{c} = 1$,so $c = 2$.
The equation of the plane is $x + y + \frac{z}{2} = 1$,or $2x + 2y + z - 2 = 0$.
The normal vector to the plane is $\vec{n} = 2\hat{i} + 2\hat{j} + 1\hat{k}$.
The direction vector of the $X$-axis is $\vec{v} = 1\hat{i} + 0\hat{j} + 0\hat{k}$.
The angle $\theta$ between a line with direction vector $\vec{v}$ and a plane with normal $\vec{n}$ is given by $\sin \theta = \frac{|\vec{n} \cdot \vec{v}|}{|\vec{n}| |\vec{v}|}$.
$\sin \theta = \frac{|(2)(1) + (2)(0) + (1)(0)|}{\sqrt{2^2 + 2^2 + 1^2} \sqrt{1^2 + 0^2 + 0^2}} = \frac{2}{\sqrt{4 + 4 + 1} \cdot 1} = \frac{2}{\sqrt{9}} = \frac{2}{3}$.
Thus,$\theta = \sin^{-1}\left(\frac{2}{3}\right)$.

(C) Since the line lies in the plane,the direction vector of the line is perpendicular to the normal vector of the plane. The direction vector of the line is $\vec{v} = 2\hat{i} - \hat{j} + 3\hat{k}$ and the normal to the plane is $\vec{n} = \ell\hat{i} + m\hat{j} - \hat{k}$.
Thus,$\vec{v} \cdot \vec{n} = 0 \Rightarrow (2)(\ell) + (-1)(m) + (3)(-1) = 0 \Rightarrow 2\ell - m = 3$ (Equation $i$).
Also,any point on the line must satisfy the plane equation. The point $(3, -2, -4)$ lies on the line,so it must lie on the plane:
$\ell(3) + m(-2) - (-4) = 9 \Rightarrow 3\ell - 2m + 4 = 9 \Rightarrow 3\ell - 2m = 5$ (Equation $ii$).
Solving the system of equations:
From $(i)$,$m = 2\ell - 3$. Substituting into $(ii)$:
$3\ell - 2(2\ell - 3) = 5 \Rightarrow 3\ell - 4\ell + 6 = 5 \Rightarrow -\ell = -1 \Rightarrow \ell = 1$.
Then $m = 2(1) - 3 = -1$.
Therefore,$\ell^2 + m^2 = (1)^2 + (-1)^2 = 1 + 1 = 2$.

(D) Let the given line be $L: \frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5} = k$. Any point on $L$ is $(3k+1, k+3, -5k+4)$.
For the intersection point of $L$ and the plane $2x-y+z+3=0$,we have $2(3k+1)-(k+3)+(-5k+4)+3=0$.
$6k+2-k-3-5k+4+3=0 \implies 6=0$,which is impossible. Thus,the line is parallel to the plane.
The image of point $P(1, 3, 4)$ on the line in the plane $2x-y+z+3=0$ is $P'(x', y', z')$.
Using the formula $\frac{x'-1}{2} = \frac{y'-3}{-1} = \frac{z'-4}{1} = -2 \frac{2(1)-3+4+3}{2^2+(-1)^2+1^2} = -2 \frac{6}{6} = -2$.
$x'-1 = -4 \implies x' = -3$.
$y'-3 = 2 \implies y' = 5$.
$z'-4 = -2 \implies z' = 2$.
The image line passes through $(-3, 5, 2)$ and has the same direction ratios $(3, 1, -5)$ as the original line.
Thus,the equation is $\frac{x+3}{3} = \frac{y-5}{1} = \frac{z-2}{-5}$.

(C) The given line is $\frac{x-4}{1} = \frac{y-2}{1} = \frac{z - m/2}{3/2}$.
This line passes through the point $P(4, 2, m/2)$.
Since the line lies in the plane $2x - 5y + 2z = 7$,the point $P$ must satisfy the equation of the plane.
Substituting the coordinates of $P$ into the plane equation:
$2(4) - 5(2) + 2(m/2) = 7$
$8 - 10 + m = 7$
$-2 + m = 7$
$m = 9$
Also,the direction vector of the line $\vec{v} = (1, 1, 3/2)$ must be perpendicular to the normal of the plane $\vec{n} = (2, -5, 2)$.
Checking the dot product: $\vec{v} \cdot \vec{n} = (1)(2) + (1)(-5) + (3/2)(2) = 2 - 5 + 3 = 0$.
Since the dot product is $0$,the line is parallel to the plane,and since the point $P$ lies on the plane,the entire line lies in the plane for $m = 9$.

(A) Let the line be $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}=\lambda$.
Any point on this line is given by $(2\lambda+4, 2\lambda+5, \lambda+3)$.
Since this point lies on the plane $x+y+z=2$,we substitute the coordinates into the plane equation:
$(2\lambda+4) + (2\lambda+5) + (\lambda+3) = 2$.
$5\lambda + 12 = 2$.
$5\lambda = -10$,so $\lambda = -2$.
Substituting $\lambda = -2$ back into the point coordinates:
$x = 2(-2) + 4 = 0$,
$y = 2(-2) + 5 = 1$,
$z = (-2) + 3 = 1$.
The point of intersection is $(0, 1, 1)$.
Now,check which option satisfies the point $(0, 1, 1)$:
For option $(A)$: $\frac{0-1}{1} = -1$,$\frac{1-3}{2} = -1$,$\frac{1+4}{-5} = -1$.
Since all ratios are equal to $-1$,the point $(0, 1, 1)$ lies on the line given in option $(A)$.

(B) Two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ are coplanar if $\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right| = 0$.
Here,$(x_1, y_1, z_1) = (2, 3, 4)$,$(a_1, b_1, c_1) = (1, 1, -k)$,$(x_2, y_2, z_2) = (1, 4, 5)$,and $(a_2, b_2, c_2) = (k, 2, 1)$.
Substituting these values into the determinant condition:
$\left|\begin{array}{ccc} 1-2 & 4-3 & 5-4 \\ 1 & 1 & -k \\ k & 2 & 1 \end{array}\right| = 0$
$\left|\begin{array}{ccc} -1 & 1 & 1 \\ 1 & 1 & -k \\ k & 2 & 1 \end{array}\right| = 0$
Expanding the determinant along the first row:
$-1(1 - (-2k)) - 1(1 - (-k^2)) + 1(2 - k) = 0$
$-1(1 + 2k) - 1(1 + k^2) + 2 - k = 0$
$-1 - 2k - 1 - k^2 + 2 - k = 0$
$-k^2 - 3k = 0$
$k^2 + 3k = 0$
$k(k + 3) = 0$
Therefore,$k = 0$ or $k = -3$.

(A) Let $A(x_1, y_1, z_1) = (3, 4, 1)$ and $B(x_2, y_2, z_2) = (5, 1, 6)$.
The equation of the line passing through two points is $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$.
Substituting the values,we get $\frac{x-3}{5-3} = \frac{y-4}{1-4} = \frac{z-1}{6-1}$,which simplifies to $\frac{x-3}{2} = \frac{y-4}{-3} = \frac{z-1}{5} = k$.
Since the line crosses the $xy$-plane,the $z$-coordinate must be $0$.
Setting $z = 0$,we have $\frac{0-1}{5} = k$,so $k = -\frac{1}{5}$.
Now,find $x$ and $y$ using $k = -\frac{1}{5}$:
$x - 3 = 2k \Rightarrow x = 3 + 2(-\frac{1}{5}) = 3 - \frac{2}{5} = \frac{13}{5}$.
$y - 4 = -3k \Rightarrow y = 4 - 3(-\frac{1}{5}) = 4 + \frac{3}{5} = \frac{23}{5}$.
Thus,the required point is $\left(\frac{13}{5}, \frac{23}{5}, 0\right)$.

(B) $1$. Direction Cosines: Since the line makes equal angles with the coordinate axes,its direction ratios are proportional to $(1, 1, 1)$.
$2$. Equation of the Line: The parametric form of the line passing through $P(2, 1, 2)$ with direction ratios $(1, 1, 1)$ is:
$\frac{x-2}{1} = \frac{y-1}{1} = \frac{z-2}{1} = t$
Thus,$x = 2+t, y = 1+t, z = 2+t$.
$3$. Point of Intersection with Plane: Substituting these into the plane equation $2x+y+z=9$:
$2(2+t) + (1+t) + (2+t) = 9$
$4 + 2t + 1 + t + 2 + t = 9$
$4t + 7 = 9 \Rightarrow 4t = 2 \Rightarrow t = \frac{1}{2}$.
$4$. Coordinates of $Q$: Substituting $t = \frac{1}{2}$ back,we get $Q = (2+\frac{1}{2}, 1+\frac{1}{2}, 2+\frac{1}{2}) = (\frac{5}{2}, \frac{3}{2}, \frac{5}{2})$.
$5$. Length of $PQ$: Using the distance formula:
$PQ = \sqrt{(\frac{5}{2}-2)^2 + (\frac{3}{2}-1)^2 + (\frac{5}{2}-2)^2}$
$PQ = \sqrt{(\frac{1}{2})^2 + (\frac{1}{2})^2 + (\frac{1}{2})^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$ units.

(B) For a line to lie in a plane,every point on the line must satisfy the equation of the plane. The line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z+m}{2}$ passes through the point $(4, 2, -m)$.
Substituting this point into the plane equation $2x - 4y + z = 7$:
$2(4) - 4(2) + (-m) = 7$
$8 - 8 - m = 7$
$-m = 7$
$m = -7$
Additionally,the normal vector of the plane $\vec{n} = (2, -4, 1)$ must be perpendicular to the direction vector of the line $\vec{v} = (1, 1, 2)$.
Checking the dot product: $(2)(1) + (-4)(1) + (1)(2) = 2 - 4 + 2 = 0$.
Since the dot product is $0$,the line is parallel to the plane. Since the point $(4, 2, -m)$ lies on the plane for $m = -7$,the entire line lies in the plane.

(B) The given equation of the line is $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$.
This line passes through the point $P(2, 1, -2)$.
Since the line lies in the plane $x+3y-\alpha z+\beta=0$,the point $P$ must satisfy the plane equation:
$2 + 3(1) - \alpha(-2) + \beta = 0$
$2 + 3 + 2\alpha + \beta = 0$
$2\alpha + \beta = -5$ $(i)$
Also,the direction vector of the line $\vec{v} = (3, -5, 2)$ is perpendicular to the normal vector of the plane $\vec{n} = (1, 3, -\alpha)$.
Therefore,their dot product is zero:
$3(1) + (-5)(3) + 2(-\alpha) = 0$
$3 - 15 - 2\alpha = 0$
$-12 - 2\alpha = 0$
$2\alpha = -12 \implies \alpha = -6$
Substituting $\alpha = -6$ into equation $(i)$:
$2(-6) + \beta = -5$
$-12 + \beta = -5$
$\beta = 7$
Thus,$(\alpha, \beta) = (-6, 7)$.

(D) Since the line lies in the plane,the direction vector of the line is perpendicular to the normal vector of the plane. The direction vector of the line is $\vec{v} = 2\hat{i} + \hat{j} + 3\hat{k}$ and the normal to the plane is $\vec{n} = \ell\hat{i} + m\hat{j} - \hat{k}$.
Thus,$\vec{v} \cdot \vec{n} = 0 \Rightarrow 2\ell + m - 3 = 0$,which gives $2\ell + m = 3$ (Equation $i$).
Also,any point on the line must lie on the plane. The point $(3, -2, -4)$ lies on the line,so it must satisfy the plane equation $\ell x + my - z = 9$.
Substituting the point: $3\ell - 2m - (-4) = 9 \Rightarrow 3\ell - 2m = 5$ (Equation $ii$).
Solving the system of equations:
From $(i)$,$m = 3 - 2\ell$. Substituting into $(ii)$:
$3\ell - 2(3 - 2\ell) = 5 \Rightarrow 3\ell - 6 + 4\ell = 5 \Rightarrow 7\ell = 11 \Rightarrow \ell = \frac{11}{7}$.
Then $m = 3 - 2(\frac{11}{7}) = \frac{21 - 22}{7} = -\frac{1}{7}$.
Finally,$\ell^2 + m^2 = (\frac{11}{7})^2 + (-\frac{1}{7})^2 = \frac{121}{49} + \frac{1}{49} = \frac{122}{49}$.

(C) The equation of the plane is $2x + 3y + 4z = 1$.
To find the intersection with the $X$-axis,set $y = 0$ and $z = 0$: $2x = 1 \implies x = \frac{1}{2}$. Thus,$A = (\frac{1}{2}, 0, 0)$.
To find the intersection with the $Y$-axis,set $x = 0$ and $z = 0$: $3y = 1 \implies y = \frac{1}{3}$. Thus,$B = (0, \frac{1}{3}, 0)$.
To find the intersection with the $Z$-axis,set $x = 0$ and $y = 0$: $4z = 1 \implies z = \frac{1}{4}$. Thus,$C = (0, 0, \frac{1}{4})$.
The centroid of $\triangle ABC$ with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$.
Substituting the coordinates:
Centroid $= (\frac{1/2 + 0 + 0}{3}, \frac{0 + 1/3 + 0}{3}, \frac{0 + 0 + 1/4}{3}) = (\frac{1}{6}, \frac{1}{9}, \frac{1}{12})$.

(A) The position vector of point $Q$ on the line is given by $\vec{q} = (1 - 3\mu)\hat{i} + (-1 + \mu)\hat{j} + (2 + 5\mu)\hat{k}$.
Given point $P$ is $(3, 2, 6)$,so its position vector is $\vec{p} = 3\hat{i} + 2\hat{j} + 6\hat{k}$.
The vector $\vec{PQ} = \vec{q} - \vec{p} = (-3\mu - 2)\hat{i} + (\mu - 3)\hat{j} + (5\mu - 4)\hat{k}$.
The normal vector to the plane $x - 4y + 3z = 1$ is $\vec{n} = \hat{i} - 4\hat{j} + 3\hat{k}$.
Since $\vec{PQ}$ is parallel to the plane,it must be perpendicular to the normal vector $\vec{n}$,so $\vec{PQ} \cdot \vec{n} = 0$.
$(-3\mu - 2)(1) + (\mu - 3)(-4) + (5\mu - 4)(3) = 0$.
$-3\mu - 2 - 4\mu + 12 + 15\mu - 12 = 0$.
$8\mu - 2 = 0$.
$8\mu = 2 \Rightarrow \mu = \frac{2}{8} = \frac{1}{4}$.

(B) The lines $L_1$ and $L_2$ are coplanar if the determinant of the vector connecting a point on each line and their direction vectors is zero.
Given points: $P_1 = (-1, 2, 1)$ and $P_2 = (-2, -1, -1)$.
Direction vectors: $\vec{v_1} = (2, -1, 1)$ and $\vec{v_2} = (\alpha, 5-\alpha, 1)$.
The condition for coplanarity is:
$\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$
$\begin{vmatrix} -2-(-1) & -1-2 & -1-1 \\ 2 & -1 & 1 \\ \alpha & 5-\alpha & 1 \end{vmatrix} = 0$
$\begin{vmatrix} -1 & -3 & -2 \\ 2 & -1 & 1 \\ \alpha & 5-\alpha & 1 \end{vmatrix} = 0$
Expanding the determinant:
$-1(-1 - (5-\alpha)) + 3(2 - \alpha) - 2(2(5-\alpha) - (-1)\alpha) = 0$
$-1(-6+\alpha) + 6 - 3\alpha - 2(10 - 2\alpha + \alpha) = 0$
$6 - \alpha + 6 - 3\alpha - 20 + 2\alpha = 0$
$-2\alpha - 8 = 0 \Rightarrow \alpha = -4$.
Substituting $\alpha = -4$ into $L_2$:
$L_2: \frac{x+2}{-4} = \frac{y+1}{5-(-4)} = \frac{z+1}{1} \Rightarrow \frac{x+2}{-4} = \frac{y+1}{9} = \frac{z+1}{1}$.
Checking option $(B) (2, -10, -2)$:
$\frac{2+2}{-4} = \frac{-10+1}{9} = \frac{-2+1}{1} \Rightarrow -1 = -1 = -1$.
Thus,the line $L_2$ passes through $(2, -10, -2)$.

(NONE) Given point $P = (2, 1, 5)$.
Any point $Q$ on the line $\vec{r} = (1, -1, 2) + \mu(-3, 1, 5)$ is given by $Q = (1-3\mu, -1+\mu, 2+5\mu)$.
The vector $\vec{PQ} = Q - P = (1-3\mu-2, -1+\mu-1, 2+5\mu-5) = (-1-3\mu, -2+\mu, -3+5\mu)$.
The normal vector to the plane $3x-y+4z=1$ is $\vec{n} = (3, -1, 4)$.
For $\vec{PQ}$ to be parallel to the plane,$\vec{PQ} \cdot \vec{n} = 0$.
$3(-1-3\mu) - 1(-2+\mu) + 4(-3+5\mu) = 0$.
$-3 - 9\mu + 2 - \mu - 12 + 20\mu = 0$.
$10\mu - 13 = 0$.
$\mu = \frac{13}{10}$.

(A) The equation of the plane containing the two lines is given by the determinant form:
$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$
Using the point $(1, 4, -4)$ from the second line and the direction vectors $(3, 5, 7)$ and $(1, 4, 7)$:
$\begin{vmatrix} x-1 & y-4 & z+4 \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{vmatrix} = 0$
Expanding the determinant:
$(x-1)(35-28) - (y-4)(21-7) + (z+4)(12-5) = 0$
$7(x-1) - 14(y-4) + 7(z+4) = 0$
Dividing by $7$:
$(x-1) - 2(y-4) + (z+4) = 0$
$x - 2y + z - 1 + 8 + 4 = 0$
$x - 2y + z + 11 = 0$
The perpendicular distance from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.
$d = \frac{|11|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{11}{\sqrt{1+4+1}} = \frac{11}{\sqrt{6}}$ units.

(A) Let the origin be $O(0, 0, 0)$ and the foot of the perpendicular be $M(2, 1, -2)$.
Since $OM$ is the normal to the plane,the normal vector $\vec{n}$ is given by the vector $\vec{OM} = (2 - 0)\hat{i} + (1 - 0)\hat{j} + (-2 - 0)\hat{k} = 2\hat{i} + \hat{j} - 2\hat{k}$.
The vector equation of a plane passing through a point with position vector $\vec{a}$ and having normal vector $\vec{n}$ is given by $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.
Here,$\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{n} = 2\hat{i} + \hat{j} - 2\hat{k}$.
Thus,$\vec{r} \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = (2\hat{i} + \hat{j} - 2\hat{k}) \cdot (2\hat{i} + \hat{j} - 2\hat{k})$.
Calculating the dot product: $(2 \times 2) + (1 \times 1) + (-2 \times -2) = 4 + 1 + 4 = 9$.
Therefore,the equation of the plane is $\vec{r} \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = 9$.

(D) Since the vector $\overline{c}$ lies in the plane of $\overline{a}$ and $\overline{b}$,the vectors $\overline{a}, \overline{b}$,and $\overline{c}$ are coplanar.
For three vectors to be coplanar,their scalar triple product must be zero: $\overline{a} \cdot (\overline{b} \times \overline{c}) = 0$.
This is equivalent to the determinant of the components being zero:
$\begin{vmatrix} 1 & -1 & 2 \\ 1 & 1 & 1 \\ x & -(2-x) & -1 \end{vmatrix} = 0$
$\Rightarrow 1(-1 - (-(2-x))) - (-1)(-1 - x) + 2(-(2-x) - x) = 0$
$\Rightarrow 1(-1 + 2 - x) + 1(-1 - x) + 2(-2 + x - x) = 0$
$\Rightarrow 1(1 - x) - 1 - x + 2(-2) = 0$
$\Rightarrow 1 - x - 1 - x - 4 = 0$
$\Rightarrow -2x - 4 = 0$
$\Rightarrow -2x = 4$
$\Rightarrow x = -2$

(D) The equation of the family of planes passing through the line of intersection of the planes $x+y+z-1=0$ and $2x+3y+4z-5=0$ is given by $(x+y+z-1) + \lambda(2x+3y+4z-5) = 0$.
Rearranging the terms,we get $(1+2\lambda)x + (1+3\lambda)y + (1+4\lambda)z - (1+5\lambda) = 0$.
This plane is perpendicular to the plane $x-y+z=0$.
The normal vectors of these two planes are $\vec{n_1} = (1+2\lambda)\hat{i} + (1+3\lambda)\hat{j} + (1+4\lambda)\hat{k}$ and $\vec{n_2} = \hat{i} - \hat{j} + \hat{k}$.
Since the planes are perpendicular,their dot product is zero: $\vec{n_1} \cdot \vec{n_2} = 0$.
$(1+2\lambda)(1) + (1+3\lambda)(-1) + (1+4\lambda)(1) = 0$.
$1+2\lambda - 1 - 3\lambda + 1 + 4\lambda = 0$.
$1 + 3\lambda = 0 \Rightarrow \lambda = -\frac{1}{3}$.
Substituting $\lambda = -\frac{1}{3}$ into the equation of the plane:
$(1 - \frac{2}{3})x + (1 - 1)y + (1 - \frac{4}{3})z - (1 - \frac{5}{3}) = 0$.
$\frac{1}{3}x + 0y - \frac{1}{3}z + \frac{2}{3} = 0$.
Multiplying by $3$,we get $x - z + 2 = 0$.
The vector equation is $\overline{r} \cdot (\hat{i} - \hat{k}) + 2 = 0$.

(D) The equation of a plane passing through $(x_0, y_0, z_0)$ is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.
Given the point $(2, -1, -3)$,the equation is $a(x-2) + b(y+1) + c(z+3) = 0$.
Since the plane is parallel to the lines with direction ratios $(3, 2, -4)$ and $(2, -3, 2)$,the normal vector $\vec{n} = (a, b, c)$ must be perpendicular to both direction vectors.
Thus,$\vec{n} = (3, 2, -4) \times (2, -3, 2) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -4 \\ 2 & -3 & 2 \end{vmatrix} = \hat{i}(4 - 12) - \hat{j}(6 + 8) + \hat{k}(-9 - 4) = -8\hat{i} - 14\hat{j} - 13\hat{k}$.
Taking the normal vector as $(8, 14, 13)$,the equation becomes $8(x-2) + 14(y+1) + 13(z+3) = 0$.
Expanding this,we get $8x - 16 + 14y + 14 + 13z + 39 = 0$,which simplifies to $8x + 14y + 13z + 37 = 0$.

(A) Given $f(x) = 3 \sin x - 4 \sin^3 x$.
Using the identity $\sin 3x = 3 \sin x - 4 \sin^3 x$,we have $f(x) = \sin 3x$.
To find the interval where $f(x)$ is increasing,we find the derivative $f'(x) = 3 \cos 3x$.
For $f(x)$ to be increasing,$f'(x) \geq 0$,which implies $3 \cos 3x \geq 0$,or $\cos 3x \geq 0$.
The cosine function is non-negative in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Thus,$-\frac{\pi}{2} \leq 3x \leq \frac{\pi}{2}$.
Dividing by $3$,we get $-\frac{\pi}{6} \leq x \leq \frac{\pi}{6}$.
The length of this interval is $\frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{2\pi}{6} = \frac{\pi}{3}$.

(C) Let $x = 2 \tan ^{-1}\left(\frac{1}{5}\right)$.
Then $\tan \left(\frac{x}{2}\right) = \frac{1}{5}$.
Using the formula $\tan x = \frac{2 \tan (x/2)}{1 - \tan^2 (x/2)}$,we get:
$\tan x = \frac{2(1/5)}{1 - (1/5)^2} = \frac{2/5}{1 - 1/25} = \frac{2/5}{24/25} = \frac{2}{5} \times \frac{25}{24} = \frac{5}{12}$.
Now,we need to evaluate $\tan \left(x + \frac{\pi}{4}\right)$.
Using the formula $\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have:
$\tan \left(x + \frac{\pi}{4}\right) = \frac{\tan x + \tan(\pi/4)}{1 - \tan x \tan(\pi/4)}$.
Substituting $\tan x = \frac{5}{12}$ and $\tan(\pi/4) = 1$:
$= \frac{5/12 + 1}{1 - (5/12)(1)} = \frac{17/12}{7/12} = \frac{17}{7}$.

(D) Given the expression $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$.
We can rewrite the argument as $\cos \left(\cos ^{-1} x + (\cos ^{-1} x + \sin ^{-1} x)\right)$.
Using the identity $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,the expression becomes $\cos \left(\cos ^{-1} x + \frac{\pi}{2}\right)$.
Using the trigonometric identity $\cos \left(\frac{\pi}{2} + \theta\right) = -\sin \theta$,we get $-\sin \left(\cos ^{-1} x\right)$.
Since $\sin \left(\cos ^{-1} x\right) = \sqrt{1-x^2}$,the expression simplifies to $-\sqrt{1-x^2}$.
Substituting $x = \frac{1}{5}$:
$-\sqrt{1-\left(\frac{1}{5}\right)^2} = -\sqrt{1-\frac{1}{25}} = -\sqrt{\frac{24}{25}} = -\frac{2 \sqrt{6}}{5}$.

(C) The given Cartesian equations of the line are $y=2$ and $4x-3z+5=0$.
We can rewrite these as $y=2$ and $4x=3z-5$.
Dividing by $12$ to get the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$:
$4x = 3(z - \frac{5}{3}) \implies \frac{x}{3} = \frac{z - 5/3}{4}$.
Since $y=2$ is a constant,the direction ratio for $y$ is $0$.
Thus,the line can be written as $\frac{x-0}{3} = \frac{y-2}{0} = \frac{z-5/3}{4}$.
The line passes through the point $(0, 2, 5/3)$ and has direction ratios $(3, 0, 4)$.
The vector equation is $\overline{r} = \vec{a} + \lambda \vec{b}$,where $\vec{a} = 0\hat{i} + 2\hat{j} + \frac{5}{3}\hat{k}$ and $\vec{b} = 3\hat{i} + 0\hat{j} + 4\hat{k}$.
Therefore,$\overline{r} = (2\hat{j} + \frac{5}{3}\hat{k}) + \lambda(3\hat{i} + 4\hat{k})$.

(C) Let $\vec{a}=5 \hat{i}+2 \hat{j}+6 \hat{k}$,$\vec{b}=2 \hat{i}+\hat{j}+\hat{k}$,and $\vec{c}=\hat{i}-\hat{j}+\hat{k}$.
Since the required vector is orthogonal to $\vec{a}$ and coplanar with $\vec{b}$ and $\vec{c}$,it must be parallel to $\vec{a} \times (\vec{b} \times \vec{c})$.
Using the vector triple product formula: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$.
Calculate the dot products:
$\vec{a} \cdot \vec{c} = (5)(1) + (2)(-1) + (6)(1) = 5 - 2 + 6 = 9$.
$\vec{a} \cdot \vec{b} = (5)(2) + (2)(1) + (6)(1) = 10 + 2 + 6 = 18$.
Substitute these into the formula:
$\vec{a} \times (\vec{b} \times \vec{c}) = 9(2 \hat{i} + \hat{j} + \hat{k}) - 18(\hat{i} - \hat{j} + \hat{k}) = (18-18)\hat{i} + (9+18)\hat{j} + (9-18)\hat{k} = 27 \hat{j} - 9 \hat{k}$.
The magnitude is $|27 \hat{j} - 9 \hat{k}| = \sqrt{27^2 + (-9)^2} = \sqrt{729 + 81} = \sqrt{810} = 9 \sqrt{10}$.
The unit vector is $\pm \frac{27 \hat{j} - 9 \hat{k}}{9 \sqrt{10}} = \pm \frac{3 \hat{j} - \hat{k}}{\sqrt{10}}$.
Comparing with the options,$\frac{3 \hat{j} - \hat{k}}{\sqrt{10}}$ is the correct choice.

(D) We know that the scalar triple product $[\overline{a} \overline{b} \overline{c}] = (\overline{a} \times \overline{b}) \cdot \overline{c}$.
For the first term: $\overline{p} \cdot (\overline{a} + \overline{b}) = \overline{p} \cdot \overline{a} + \overline{p} \cdot \overline{b}$.
Substituting $\overline{p} = \frac{\overline{b} \times \overline{c}}{[\overline{a} \overline{b} \overline{c}]}$,we get:
$\overline{p} \cdot \overline{a} = \frac{(\overline{b} \times \overline{c}) \cdot \overline{a}}{[\overline{a} \overline{b} \overline{c}]} = \frac{[\overline{b} \overline{c} \overline{a}]}{[\overline{a} \overline{b} \overline{c}]} = \frac{[\overline{a} \overline{b} \overline{c}]}{[\overline{a} \overline{b} \overline{c}]} = 1$.
$\overline{p} \cdot \overline{b} = \frac{(\overline{b} \times \overline{c}) \cdot \overline{b}}{[\overline{a} \overline{b} \overline{c}]} = 0$ (since $\overline{b} \times \overline{c}$ is perpendicular to $\overline{b}$).
Thus,$\overline{p} \cdot (\overline{a} + \overline{b}) = 1 + 0 = 1$.
Similarly,$\overline{q} \cdot (\overline{b} + \overline{c}) = 1$ and $\overline{r} \cdot (\overline{c} + \overline{a}) = 1$.
Adding these,we get $1 + 1 + 1 = 3$.

(C) Let $\overline{d} = p\hat{i} + q\hat{j} + r\hat{k}$,where $p, q, r \in \mathbb{R}$.
Since $\overline{b}, \overline{c}, \overline{d}$ are coplanar,their scalar triple product is zero:
$\begin{vmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ p & q & r \end{vmatrix} = 0$
Expanding the determinant: $0(0 - q) - 1(-r - p) + (-1)(-q - 0) = 0$
$r + p + q = 0 \implies p + q + r = 0 \dots (i)$
Given that $\overline{a}$ and $\overline{d}$ are perpendicular,their dot product is zero:
$\overline{a} \cdot \overline{d} = (1, -1, 0) \cdot (p, q, r) = p - q = 0 \implies p = q \dots (ii)$
Substituting $(ii)$ into $(i)$: $p + p + r = 0 \implies r = -2p$.
Thus,$\overline{d} = p\hat{i} + p\hat{j} - 2p\hat{k} = p(\hat{i} + \hat{j} - 2\hat{k})$.
Since $\overline{d}$ is a unit vector,$|\overline{d}| = 1$:
$|p|\sqrt{1^2 + 1^2 + (-2)^2} = 1 \implies |p|\sqrt{6} = 1 \implies p = \pm \frac{1}{\sqrt{6}}$.
For $p = \frac{1}{\sqrt{6}}$,$\overline{d} = \frac{1}{\sqrt{6}}(1, 1, -2)$,which matches option $(C)$.

(A) Given $\overline{a} \times \overline{b} + \overline{c} = \overline{0}$,which implies $\overline{a} \times \overline{b} = -\overline{c}$.
Taking the cross product with $\overline{a}$ on both sides: $\overline{a} \times (\overline{a} \times \overline{b}) = -\overline{a} \times \overline{c}$.
Using the vector triple product formula $\overline{a} \times (\overline{b} \times \overline{c}) = (\overline{a} \cdot \overline{c})\overline{b} - (\overline{a} \cdot \overline{b})\overline{c}$,we get $(\overline{a} \cdot \overline{b})\overline{a} - (\overline{a} \cdot \overline{a})\overline{b} = -\overline{a} \times \overline{c}$.
Given $\overline{a} = \hat{j} - \hat{k}$,so $\overline{a} \cdot \overline{a} = 0^2 + 1^2 + (-1)^2 = 2$.
Given $\overline{a} \cdot \overline{b} = 3$,so $3\overline{a} - 2\overline{b} = -\overline{a} \times \overline{c}$.
Calculate $\overline{a} \times \overline{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & -1 \\ 1 & -1 & -1 \end{vmatrix} = \hat{i}(-1 - 1) - \hat{j}(0 - (-1)) + \hat{k}(0 - 1) = -2\hat{i} - \hat{j} - \hat{k}$.
Thus,$2\overline{b} = 3\overline{a} + (\overline{a} \times \overline{c}) = 3(\hat{j} - \hat{k}) + (-2\hat{i} - \hat{j} - \hat{k}) = -2\hat{i} + 2\hat{j} - 4\hat{k}$.
Therefore,$\overline{b} = -\hat{i} + \hat{j} - 2\hat{k}$.

(C) The lines $L_1$ and $L_2$ are parallel to the vectors $\vec{b}_1 = 3 \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{b}_2 = 2 \hat{i} + \hat{j} + 3 \hat{k}$ respectively.
The unit vector perpendicular to both $L_1$ and $L_2$ is given by $\hat{n} = \frac{\vec{b}_1 \times \vec{b}_2}{|\vec{b}_1 \times \vec{b}_2|}$.
First,we calculate the cross product $\vec{b}_1 \times \vec{b}_2$:
$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{vmatrix} = \hat{i}(6-1) - \hat{j}(9-2) + \hat{k}(3-4) = 5 \hat{i} - 7 \hat{j} - \hat{k}$.
Next,we find the magnitude of the cross product:
$|\vec{b}_1 \times \vec{b}_2| = \sqrt{5^2 + (-7)^2 + (-1)^2} = \sqrt{25 + 49 + 1} = \sqrt{75} = 5 \sqrt{3}$.
Therefore,the required unit vector is $\hat{n} = \frac{5 \hat{i} - 7 \hat{j} - \hat{k}}{5 \sqrt{3}}$.

(B) Given the scalar triple product equation: $[m\overline{a}+\overline{b} \quad m\overline{b}+\overline{c} \quad m\overline{c}+\overline{a}] = 28[\overline{a} \quad \overline{b} \quad \overline{c}]$.
Expanding the scalar triple product using the property $[\overline{u}+\overline{v} \quad \overline{w} \quad \overline{x}] = [\overline{u} \quad \overline{w} \quad \overline{x}] + [\overline{v} \quad \overline{w} \quad \overline{x}]$,we get:
$[m\overline{a}+\overline{b} \quad m\overline{b}+\overline{c} \quad m\overline{c}+\overline{a}] = m^3[\overline{a} \quad \overline{b} \quad \overline{c}] + [\overline{b} \quad \overline{c} \quad \overline{a}] = m^3[\overline{a} \quad \overline{b} \quad \overline{c}] + [\overline{a} \quad \overline{b} \quad \overline{c}]$.
Since $[\overline{b} \quad \overline{c} \quad \overline{a}] = [\overline{a} \quad \overline{b} \quad \overline{c}]$ by cyclic permutation.
Thus,$(m^3+1)[\overline{a} \quad \overline{b} \quad \overline{c}] = 28[\overline{a} \quad \overline{b} \quad \overline{c}]$.
Since the vectors are non-coplanar,$[\overline{a} \quad \overline{b} \quad \overline{c}] \neq 0$.
Therefore,$m^3+1 = 28 \implies m^3 = 27 \implies m = 3$.

(B) Given that $\overline{a}$ and $\overline{b}$ are not perpendicular,so $\overline{a} \cdot \overline{b} \neq 0$.
We are given $\overline{a} \cdot \overline{d} = 0$ and $\overline{b} \times \overline{c} = \overline{b} \times \overline{d}$.
Taking the cross product with $\overline{a}$ on both sides of the equation $\overline{b} \times \overline{c} = \overline{b} \times \overline{d}$,we get:
$\overline{a} \times (\overline{b} \times \overline{c}) = \overline{a} \times (\overline{b} \times \overline{d})$
Using the vector triple product formula $\overline{a} \times (\overline{b} \times \overline{c}) = (\overline{a} \cdot \overline{c}) \overline{b} - (\overline{a} \cdot \overline{b}) \overline{c}$,we have:
$(\overline{a} \cdot \overline{c}) \overline{b} - (\overline{a} \cdot \overline{b}) \overline{c} = (\overline{a} \cdot \overline{d}) \overline{b} - (\overline{a} \cdot \overline{b}) \overline{d}$
Since $\overline{a} \cdot \overline{d} = 0$,the equation simplifies to:
$(\overline{a} \cdot \overline{c}) \overline{b} - (\overline{a} \cdot \overline{b}) \overline{c} = 0 - (\overline{a} \cdot \overline{b}) \overline{d}$
Rearranging to solve for $\overline{d}$:
$(\overline{a} \cdot \overline{b}) \overline{d} = (\overline{a} \cdot \overline{b}) \overline{c} - (\overline{a} \cdot \overline{c}) \overline{b}$
Dividing by $(\overline{a} \cdot \overline{b})$:
$\overline{d} = \overline{c} - \left(\frac{\overline{a} \cdot \overline{c}}{\overline{a} \cdot \overline{b}}\right) \overline{b}$

(B) The volume $V$ of a parallelepiped formed by vectors $\vec{u}, \vec{v}, \vec{w}$ is given by the scalar triple product $|\vec{u} \cdot (\vec{v} \times \vec{w})|$.
$V = \begin{vmatrix} 1 & a & 1 \\ 0 & 1 & a \\ a & 0 & 1 \end{vmatrix} = 1(1 - 0) - a(0 - a^2) + 1(0 - a) = 1 + a^3 - a$.
To find the minimum volume,we differentiate $V$ with respect to $a$:
$\frac{dV}{da} = 3a^2 - 1$.
Setting $\frac{dV}{da} = 0$,we get $3a^2 = 1$,so $a^2 = \frac{1}{3}$,which implies $a = \pm \frac{1}{\sqrt{3}}$.
Now,we check the second derivative: $\frac{d^2V}{da^2} = 6a$.
For $a = \frac{1}{\sqrt{3}}$,$\frac{d^2V}{da^2} = 6(\frac{1}{\sqrt{3}}) > 0$,which indicates a local minimum.
For $a = -\frac{1}{\sqrt{3}}$,$\frac{d^2V}{da^2} = 6(-\frac{1}{\sqrt{3}}) < 0$,which indicates a local maximum.
Thus,the volume is minimum at $a = \frac{1}{\sqrt{3}}$.

(A) Let $\vec{r}$ and $\vec{s}$ be the position vectors of points $R$ and $S$ respectively.
Using the section formula for internal division in ratio $2:3$:
$\vec{r} = \frac{2\vec{q} + 3\vec{p}}{2+3} = \frac{3\vec{p} + 2\vec{q}}{5}$
Using the section formula for external division in ratio $2:3$:
$\vec{s} = \frac{2\vec{q} - 3\vec{p}}{2-3} = \frac{2\vec{q} - 3\vec{p}}{-1} = 3\vec{p} - 2\vec{q}$
Since $\vec{OR}$ and $\vec{OS}$ are perpendicular,their dot product is zero:
$\vec{r} \cdot \vec{s} = 0$
$\left(\frac{3\vec{p} + 2\vec{q}}{5}\right) \cdot (3\vec{p} - 2\vec{q}) = 0$
$(3\vec{p} + 2\vec{q}) \cdot (3\vec{p} - 2\vec{q}) = 0$
$9|\vec{p}|^2 - 6(\vec{p} \cdot \vec{q}) + 6(\vec{q} \cdot \vec{p}) - 4|\vec{q}|^2 = 0$
$9p^2 - 4q^2 = 0$
$9p^2 = 4q^2$

(C) The projection of $\overline{a}$ on $\overline{c}$ is given by $\frac{\overline{a} \cdot \overline{c}}{|\overline{c}|}$.
Given $\overline{a} = \alpha \hat{i} + 3 \hat{j} - \hat{k}$ and $\overline{c} = \hat{i} + 2 \hat{j} - 2 \hat{k}$.
$\overline{a} \cdot \overline{c} = (\alpha)(1) + (3)(2) + (-1)(-2) = \alpha + 6 + 2 = \alpha + 8$.
$|\overline{c}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
So,$\frac{\alpha + 8}{3} = \frac{10}{3} \implies \alpha + 8 = 10 \implies \alpha = 2$.
Now,$\overline{b} \times \overline{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -\beta & 4 \\ 1 & 2 & -2 \end{vmatrix} = \hat{i}(2\beta - 8) - \hat{j}(-6 - 4) + \hat{k}(6 + \beta) = (2\beta - 8)\hat{i} + 10\hat{j} + (6 + \beta)\hat{k}$.
Comparing this with $-6\hat{i} + 10\hat{j} + 7\hat{k}$,we get $2\beta - 8 = -6 \implies 2\beta = 2 \implies \beta = 1$.
Finally,$2\alpha + \beta = 2(2) + 1 = 4 + 1 = 5$.

(C) Given that $\bar{s} = 4\bar{p} + 3\bar{q} + 5\bar{r}$.
Also,$\bar{s} = x(-\bar{p} + \bar{q} + \bar{r}) + y(\bar{p} - \bar{q} + \bar{r}) + z(-\bar{p} - \bar{q} + \bar{r})$.
Expanding the right side,we get $\bar{s} = (-x + y - z)\bar{p} + (x - y - z)\bar{q} + (x + y + z)\bar{r}$.
Comparing the coefficients of $\bar{p}, \bar{q}, \bar{r}$ with the given expression for $\bar{s}$,we have:
$-x + y - z = 4$ $(i)$
$x - y - z = 3$ $(ii)$
$x + y + z = 5$ $(iii)$
Adding $(ii)$ and $(iii)$,we get $2x = 8 \implies x = 4$.
Adding $(i)$ and $(ii)$,we get $-2z = 7 \implies z = -\frac{7}{2}$.
Substituting $x=4$ and $z=-\frac{7}{2}$ into $(iii)$,we get $4 + y - \frac{7}{2} = 5 \implies y = 1 + \frac{7}{2} = \frac{9}{2}$.
Now,calculate $2x + y + z = 2(4) + \frac{9}{2} - \frac{7}{2} = 8 + \frac{2}{2} = 8 + 1 = 9$.

(A) Since the vectors are coplanar,their scalar triple product is zero: $\left|\begin{array}{lll}a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c\end{array}\right|=0$.
Applying $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$,we get $\left|\begin{array}{ccc} a & 1 & 1 \\ 1-a & b-1 & 0 \\ 1-a & 0 & c-1 \end{array}\right|=0$.
Expanding along the first row: $a(b-1)(c-1) - 1(1-a)(c-1) + 1(0 - (b-1)(1-a)) = 0$.
$a(b-1)(c-1) + (1-a)(c-1) + (1-a)(b-1) = 0$.
Dividing the entire equation by $(1-a)(1-b)(1-c)$,we get: $\frac{a(b-1)(c-1)}{(1-a)(1-b)(1-c)} + \frac{(1-a)(c-1)}{(1-a)(1-b)(1-c)} + \frac{(1-a)(b-1)}{(1-a)(1-b)(1-c)} = 0$.
This simplifies to: $\frac{-a}{(1-a)} + \frac{1}{(1-b)} + \frac{1}{(1-c)} = 0$.
We can write $\frac{-a}{1-a}$ as $\frac{1-a-1}{1-a} = 1 - \frac{1}{1-a}$.
Substituting this back: $1 - \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 0$.
Therefore,$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$.

(A) Let $\vec{a} = \hat{i}+\hat{j}-\hat{k}$,$\vec{b} = \hat{i}+\hat{j}+\hat{k}$,and $\vec{c} = 2\hat{i}+\hat{j}+\hat{k}$.
Since the required vector is coplanar with $\vec{b}$ and $\vec{c}$,it must be of the form $\vec{v} = \vec{a} \times (\vec{b} \times \vec{c})$.
Using the vector triple product formula,$\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.
First,calculate the dot products:
$\vec{a} \cdot \vec{c} = (1)(2) + (1)(1) + (-1)(1) = 2 + 1 - 1 = 2$.
$\vec{a} \cdot \vec{b} = (1)(1) + (1)(1) + (-1)(1) = 1 + 1 - 1 = 1$.
Now,substitute these into the formula:
$\vec{v} = 2(\hat{i}+\hat{j}+\hat{k}) - 1(2\hat{i}+\hat{j}+\hat{k}) = (2\hat{i}+2\hat{j}+2\hat{k}) - (2\hat{i}+\hat{j}+\hat{k}) = \hat{j}+\hat{k}$.
The magnitude of this vector is $|\vec{v}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$.
The unit vector is $\pm \frac{\vec{v}}{|\vec{v}|} = \pm \frac{\hat{j}+\hat{k}}{\sqrt{2}}$.
Comparing with the given options,the correct choice is $A$ (considering the negative sign).

(A) Since $\overline{a}$ bisects the angle between $\overline{b}$ and $\overline{c}$,it must be proportional to the sum of the unit vectors along $\overline{b}$ and $\overline{c}$.
Let $\hat{b} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}$ and $\hat{c} = \frac{\hat{j}+\hat{k}}{\sqrt{2}}$.
The vector $\overline{a}$ is given by $\overline{a} = k(\hat{b} + \hat{c})$ for some scalar $k$.
$\overline{a} = k \left( \frac{\hat{i}+\hat{j}}{\sqrt{2}} + \frac{\hat{j}+\hat{k}}{\sqrt{2}} \right) = \frac{k}{\sqrt{2}} (\hat{i} + 2\hat{j} + \hat{k})$.
Given $\overline{a} = \alpha \hat{i} + 2 \hat{j} + \beta \hat{k}$.
Comparing the components,we have $\alpha = \frac{k}{\sqrt{2}}$,$2 = \frac{2k}{\sqrt{2}}$,and $\beta = \frac{k}{\sqrt{2}}$.
From $2 = \frac{2k}{\sqrt{2}}$,we get $k = \sqrt{2}$.
Substituting $k = \sqrt{2}$ into the expressions for $\alpha$ and $\beta$,we get $\alpha = \frac{\sqrt{2}}{\sqrt{2}} = 1$ and $\beta = \frac{\sqrt{2}}{\sqrt{2}} = 1$.
Thus,$\alpha=1$ and $\beta=1$.

(D) Given: $(\overline{a} \times \overline{b}) \times \overline{c} = \frac{1}{3}|\overline{b}||\overline{c}| \overline{a}$.
Using the vector triple product formula: $(\overline{a} \times \overline{b}) \times \overline{c} = (\overline{a} \cdot \overline{c}) \overline{b} - (\overline{b} \cdot \overline{c}) \overline{a}$.
Since no two vectors are collinear,$\overline{a}$ and $\overline{b}$ are linearly independent. Thus,the coefficient of $\overline{b}$ must be zero,so $(\overline{a} \cdot \overline{c}) = 0$.
Comparing the coefficients of $\overline{a}$,we get: $-(\overline{b} \cdot \overline{c}) = \frac{1}{3}|\overline{b}||\overline{c}|$.
Substituting the definition of the dot product,$-|\overline{b}||\overline{c}| \cos \theta = \frac{1}{3}|\overline{b}||\overline{c}|$.
Since the vectors are non-zero,we divide by $|\overline{b}||\overline{c}|$: $\cos \theta = -\frac{1}{3}$.
Using $\sin^2 \theta = 1 - \cos^2 \theta$,we have $\sin^2 \theta = 1 - (-\frac{1}{3})^2 = 1 - \frac{1}{9} = \frac{8}{9}$.
Thus,$\sin \theta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.
Therefore,$\operatorname{cosec} \theta = \frac{1}{\sin \theta} = \frac{3}{2\sqrt{2}}$.