If the plane $2 x + y - 5 z = 0$ is rotated about its line of intersection with the plane $3 x - y + 4 z - 7 = 0$ by an angle of $\frac { \pi } { 2 }$, then the plane after the rotation passes through the point
(1) $( 2 , - 2,0 )$
(2) $( - 2,2,0 )$
(3) $( 1,0,2 )$
(4) $( - 1,0 , - 2 )$

The shortest distance between the lines $\frac{x+7}{-6} = \frac{y-6}{7} = z$ and $\frac{7-x}{2} = y-2 = z-6$ is
(1) $2\sqrt{29}$
(2) 1
(3) $\sqrt{\frac{37}{2}}$
(4) (truncated)

Let $Q$ be the foot of perpendicular drawn from the point $P(1, 2, 3)$ to the plane $x + 2y + z = 14$. If $R$ is a point on the plane such that $\angle PRQ = 60^\circ$, then the area of $\triangle PQR$ is equal to

Let the plane $2 x + 3 y + z + 20 = 0$ be rotated through a right angle about its line of intersection with the plane $x - 3 y + 5 z = 8$. If the mirror image of the point $\left( 2 , - \frac { 1 } { 2 } , 2 \right)$ in the rotated plane is $B ( a , b , c )$, then
(1) $\frac { a } { 8 } = \frac { b } { 5 } = \frac { c } { - 4 }$
(2) $\frac { a } { 4 } = \frac { b } { 5 } = \frac { c } { - 2 }$
(3) $\frac { a } { 8 } = \frac { b } { - 5 } = \frac { c } { 4 }$
(4) $\frac { a } { 4 } = \frac { b } { 5 } = \frac { c } { 2 }$

If the lines $\vec { r } = ( \hat { i } - \hat { j } + \widehat { k } ) + \lambda ( 3 \hat { j } - \widehat { k } )$ and $\vec { r } = ( \alpha \hat { i } - \hat { j } ) + \mu ( 2 \hat { i } - 3 \widehat { k } )$ are co-planar, then the distance of the plane containing these two lines from the point $( \alpha , 0,0 )$ is
(1) $\frac { 2 } { 9 }$
(2) $\frac { 2 } { 11 }$
(3) $\frac { 4 } { 11 }$
(4) 2

Let $Q$ and $R$ be two points on the line $\frac { x + 1 } { 2 } = \frac { y + 2 } { 3 } = \frac { z - 1 } { 2 }$ at a distance $\sqrt { 26 }$ from the point $P ( 4,2,7 )$. Then the square of the area of the triangle $PQR$ is $\_\_\_\_$.

Let $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$, $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{c} = \hat{i} + \hat{j} - \hat{k}$. A vector in the plane of $\vec{a}$ and $\vec{b}$ whose projection on $\vec{c}$ is $\frac{1}{\sqrt{3}}$ is
(1) $4\hat{i} - \hat{j} + 4\hat{k}$
(2) $3\hat{i} + \hat{j} - 3\hat{k}$
(3) $2\hat{i} + \hat{j} - 2\hat{k}$
(4) $4\hat{i} + \hat{j} - 4\hat{k}$

Let for a triangle $ABC$ $$\begin{aligned} & \overrightarrow { AB } = - 2 \hat { i } + \hat { j } + 3 \hat { k } \\ & \overrightarrow { CB } = \alpha \hat { i } + \beta \hat { j } + \gamma \hat { k } \\ & \overrightarrow { CA } = 4 \hat { i } + 3 \hat { j } + \delta \hat { k } \end{aligned}$$ If $\delta > 0$ and the area of the triangle $ABC$ is $5 \sqrt { 6 }$ then $\overrightarrow { CB } \cdot \overrightarrow { CA }$ is equal to
(1) 60
(2) 54
(3) 108
(4) 120

Let $S$ be the set of all $( \lambda , \mu )$ for which the vectors $\lambda \hat { i } - \hat { j } + \widehat { k } , \hat { i } + 2 \hat { j } + \mu \widehat { k }$ and $3 \hat { i } - 4 \hat { j } + 5 \widehat { k }$, where $\lambda - \mu = 5$, are coplanar, then $\sum _ { ( \lambda , \mu ) \in S } 80 \left( \lambda ^ { 2 } + \mu ^ { 2 } \right)$ is equal to
(1) 2210
(2) 2130
(3) 2290
(4) 2370

Let two vertices of a triangle $ABC$ be $(2, 4, 6)$ and $(0, -2, -5)$, and its centroid be $(2, 1, -1)$. If the image of the third vertex in the plane $x + 2y + 4z = 11$ is $(\alpha, \beta, \gamma)$, then $\alpha\beta + \beta\gamma + \gamma\alpha$ is equal to
(1) 70
(2) 76
(3) 74
(4) 72

Let the image of the point $P(2, -1, 3)$ in the plane $x + 2y - z = 0$ be $Q$. Then the distance of the plane $3x + 2y + z + 29 = 0$ from the point $Q$ is
(1) $\frac{22\sqrt{2}}{7}$
(2) $\frac{24\sqrt{2}}{7}$
(3) $2\sqrt{14}$
(4) $3\sqrt{14}$

Let $\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$, and $\vec{b}$ and $\vec{c}$ be two nonzero vectors such that $|\vec{a} + \vec{b} + \vec{c}| = |\vec{a} + \vec{b} - \vec{c}|$ and $\vec{b} \cdot \vec{c} = 0$. Consider the following two statements: $A$: $|\vec{a} + \lambda\vec{c}| \geq |\vec{a}|$ for all $\lambda \in \mathbb{R}$. $B$: $\vec{a}$ and $\vec{c}$ are always parallel.
(1) only (B) is correct
(2) neither (A) nor (B) is correct
(3) only (A) is correct
(4) both (A) and (B) are correct.

The shortest distance between the lines $\frac{x+2}{1} = \frac{y}{-2} = \frac{z-5}{2}$ and $\frac{x-4}{1} = \frac{y-1}{2} = \frac{z+3}{0}$ is
(1) 8
(2) 6
(3) 7
(4) 9

The shortest distance between the lines $\frac{x-5}{1} = \frac{y-2}{2} = \frac{z-4}{-3}$ and $\frac{x+3}{1} = \frac{y+5}{4} = \frac{z-1}{-5}$ is
(1) $7\sqrt{3}$
(2) $5\sqrt{3}$
(3) $6\sqrt{3}$
(4) $4\sqrt{3}$

Let the shortest distance between the lines $L: \frac{x-5}{-2} = \frac{y-\lambda}{0} = \frac{z+\lambda}{1}$, $\lambda \geq 0$ and $L_1: x+1 = y-1 = 4-z$ be $2\sqrt{6}$. If $(\alpha, \beta, \gamma)$ lies on $L$, then which of the following is NOT possible?
(1) $\alpha + 2\gamma = 24$
(2) $2\alpha + \gamma = 7$
(3) $2\alpha - \gamma = 9$
(4) $\alpha - 2\gamma = 19$

Let $P$ be the point of intersection of the line $\frac{x+3}{3} = \frac{y+2}{1} = \frac{1-z}{2}$ and the plane $x + y + z = 2$. If the distance of the point $P$ from the plane $3x - 4y + 12z = 32$ is $q$, then $q$ and $2q$ are the roots of the equation
(1) $x^2 - 18x - 72 = 0$
(2) $x^2 - 18x + 72 = 0$
(3) $x^2 + 18x + 72 = 0$
(4) $x^2 + 18x - 72 = 0$

If the four points, whose position vectors are $3 \hat { i } - 4 \hat { j } + 2 \widehat { k } , \hat { i } + 2 \hat { j } - \widehat { k } , - 2 \hat { i } - \hat { j } + 3 \widehat { k }$ and $5 \hat { i } - 2 \alpha \hat { j } + 4 \widehat { k }$ are coplanar, then $\alpha$ is equal to
(1) $\frac { 73 } { 17 }$
(2) $- \frac { 107 } { 17 }$
(3) $- \frac { 73 } { 17 }$
(4) $\frac { 107 } { 17 }$

Let $\vec { a } = - \hat { i } - \hat { j } + \hat { k } , \vec { a } \cdot \vec { b } = 1$ and $\vec { a } \times \vec { b } = \hat { i } - \hat { j }$. Then $\vec { a } - 6 \vec { b }$ is equal to
(1) $3 ( \hat { i } - \hat { j } - \widehat { k } )$
(2) $3 ( \hat { i } + \hat { j } + \hat { k } )$
(3) $3 ( \hat { i } - \hat { j } + \widehat { k } )$
(4) $3 ( \hat { i } + \hat { j } - \widehat { k } )$

If the vectors $\vec { a } = \lambda \hat { i } + \mu \hat { j } + 4 \widehat { k } , \vec { b } = - 2 \hat { i } + 4 \hat { j } - 2 \widehat { k }$ and $\vec { c } = 2 \hat { i } + 3 \hat { j } + \widehat { k }$ are coplanar and the projection of $\vec { a }$ on the vector $\vec { b }$ is $\sqrt { 54 }$ units, then the sum of all possible values of $\lambda + \mu$ is equal to
(1) 0
(2) 6
(3) 24
(4) 18

Let $\lambda \in \mathbb{R}$, $\vec{a} = \lambda\hat{i} + 2\hat{j} - 3\hat{k}$, $\vec{b} = \hat{i} - \lambda\hat{j} + 2\hat{k}$. If $((\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b})) \times (\vec{a} - \vec{b}) = 8\hat{i} - 40\hat{j} - 24\hat{k}$, then $|\lambda(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})|^{2}$ is equal to
(1) 140
(2) 132
(3) 144
(4) 136

If the points with position vectors $\alpha \hat { \mathrm { i } } + 10 \hat { \mathrm { j } } + 13 \hat { \mathrm { k } } , 6 \hat { \mathrm { i } } + 11 \hat { \mathrm { j } } + 11 \hat { \mathrm { k } } , \frac { 9 } { 2 } \hat { \mathrm { i } } + \beta \hat { \mathrm { j } } - 8 \hat { \mathrm { k } }$ are collinear, then $( 19 \alpha - 6 \beta ) ^ { 2 }$ is equal to
(1) 36
(2) 25
(3) 49
(4) 16

Let $\vec { a } , \vec { b }$ and $\vec { c }$ be three non-zero non-coplanar vectors. Let the position vectors of four points $A , B , C$ and $D$ be $\overrightarrow { \mathrm { a } } - \overrightarrow { \mathrm { b } } + \overrightarrow { \mathrm { c } } , \lambda \overrightarrow { \mathrm { a } } - 3 \overrightarrow { \mathrm {~b} } + 4 \overrightarrow { \mathrm { c } } , - \vec { a } + 2 \vec { b } - 3 \vec { c }$ and $2 \vec { a } - 4 \vec { b } + 6 \vec { c }$ respectively. If $\overrightarrow { A B } , \overrightarrow { A C }$ and $\overrightarrow { A D }$ are coplanar, then $\lambda$ is :

Let $\vec{a}$ and $\vec{b}$ be two vectors. Let $|\vec{a}| = 1$, $|\vec{b}| = 4$ and $\vec{a} \cdot \vec{b} = 2$. If $\vec{c} = (2\vec{a} \times \vec{b}) - 3\vec{b}$, then the value of $\vec{b} \cdot \vec{c}$ is
(1) $-24$
(2) $-48$
(3) $-84$
(4) $-60$

Let $\vec { a } = 6 \hat { i } + 9 \hat { j } + 12 \hat { k } , \vec { b } = \alpha \hat { i } + 11 \hat { j } - 2 \hat { k }$ and $\vec { c }$ be vectors such that $\vec { a } \times \vec { c } = \vec { a } \times \vec { b }$. If $\vec { a } \cdot \vec { c } = - 12$, and $\vec { c } \cdot ( \hat { i } - 2 \hat { j } + \hat { k } ) = 5$ then $\vec { c } \cdot ( \hat { i } + \hat { j } + \hat { k } )$ is equal to $\_\_\_\_$

The distance of the point $P ( 4,6 , - 2 )$ from the line passing through the point $( - 3,2,3 )$ and parallel to a line with direction ratios $3,3 , - 1$ is equal to:
(1) 3
(2) $\sqrt { 6 }$
(3) $2 \sqrt { 3 }$
(4) $\sqrt { 14 }$