Let $\alpha$ be the angle between the lines whose direction cosines satisfy the equations $l + m - n = 0$ and $l ^ { 2 } + m ^ { 2 } - n ^ { 2 } = 0$. Then the value of $\sin ^ { 4 } \alpha + \cos ^ { 4 } \alpha$ is :
(1) $\frac { 5 } { 8 }$
(2) $\frac { 1 } { 2 }$
(3) $\frac { 3 } { 8 }$
(4) $\frac { 3 } { 4 }$

If for $a > 0$, the feet of perpendiculars from the points $A ( a , - 2 a , 3 )$ and $B ( 0,4,5 )$ on the plane $l x + m y + n z = 0$ are points $C ( 0 , - a , - 1 )$ and $D$ respectively, then the length of line segment $C D$ is equal to :
(1) $\sqrt { 31 }$
(2) $\sqrt { 41 }$
(3) $\sqrt { 55 }$
(4) $\sqrt { 66 }$

Let $\vec { a } = \hat { i } + \hat { j } + \hat { k }$ and $\vec { b } = \hat { j } - \hat { k }$. If $\vec { c }$ is a vector such that $\vec { a } \times \vec { c } = \vec { b }$ and $\vec { a } \cdot \vec { c } = 3$, then $\vec { a } \cdot ( \vec { b } \times \vec { c } )$ is equal to

Let the position vectors of two points $P$ and $Q$ be $3 \hat { \mathrm { i } } - \hat { \mathrm { j } } + 2 \widehat { \mathrm { k } }$ and $\hat { \mathrm { i } } + 2 \hat { \mathrm { j } } - 4 \widehat { \mathrm { k } }$, respectively. Let $R$ and $S$ be two points such that the direction ratios of lines $P R$ and $Q S$ are $( 4 , - 1,2 )$ and $( - 2,1 , - 2 )$, respectively. Let lines $P R$ and $Q S$ intersect at $T$. If the vector $\overrightarrow { T A }$ is perpendicular to both $\overrightarrow { P R }$ and $\overrightarrow { Q S }$ and the length of vector $\overrightarrow { T A }$ is $\sqrt { 5 }$ units, then the modulus of a position vector of $A$ is :
(1) $\sqrt { 482 }$
(2) $\sqrt { 171 }$
(3) $\sqrt { 5 }$
(4) $\sqrt { 227 }$

Let $P$ be a plane $l x + m y + n z = 0$ containing the line, $\frac { 1 - x } { 1 } = \frac { y + 4 } { 2 } = \frac { z + 2 } { 3 }$. If plane $P$ divides the line segment $A B$ joining points $A ( - 3 , - 6,1 )$ and $B ( 2,4 , - 3 )$ in ratio $k : 1$ then the value of $k$ is equal to :
(1) 1.5
(2) 3
(3) 2
(4) 4

Let $\overrightarrow { \mathrm { a } } = \hat { \mathrm { i } } + \alpha \hat { \mathrm { j } } + 3 \hat { \mathrm { k } }$ and $\overrightarrow { \mathrm { b } } = 3 \hat { \mathrm { i } } - \alpha \hat { \mathrm { j } } + \hat { \mathrm { k } }$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\vec { a }$ and $\vec { b }$ is $8 \sqrt { 3 }$ square units, then $\vec { a } \cdot \vec { b }$ is equal to $\underline{\hspace{1cm}}$.

Let $\vec { x }$ be a vector in the plane containing vectors $\vec { a } = 2 \hat { i } - \hat { j } + \hat { k }$ and $\vec { b } = \hat { i } + 2 \hat { j } - \hat { k }$. If the vector $\vec { x }$ is perpendicular to $( 3 \hat { i } + 2 \hat { j } - \widehat { k } )$ and its projection on $\vec { a }$ is $\frac { 17 \sqrt { 6 } } { 2 }$, then the value of $| \vec { x } | ^ { 2 }$ is equal to $\_\_\_\_$ .

Let $S$ be the mirror image of the point $Q ( 1,3,4 )$ with respect to the plane $2 x - y + z + 3 = 0$ and let $R ( 3,5 , \gamma )$ be a point of this plane. Then the square of the length of the line segment $S R$ is

Let $\vec { a } = \hat { i } + 2 \hat { j } - \widehat { k } , \vec { b } = \hat { i } - \hat { j }$ and $\vec { c } = \hat { i } - \hat { j } - \hat { k }$ be three given vectors. If $\vec { r }$ is a vector such that $\vec { r } \times \vec { a } = \vec { c } \times \vec { a }$ and $\vec { r } \cdot \vec { b } = 0$, then $\vec { r } \cdot \vec { a }$ is equal to

A line $l$ passing through origin is perpendicular to the lines $l _ { 1 } : \vec { r } = ( 3 + t ) \hat { \mathrm { i } } + ( - 1 + 2 t ) \hat { \mathrm { j } } + ( 4 + 2 t ) \hat { \mathrm { k } }$ $l _ { 2 } : \vec { r } = ( 3 + 2 s ) \hat { \mathrm { i } } + ( 3 + 2 s ) \hat { \mathrm { j } } + ( 2 + s ) \hat { \mathrm { k } }$ If the co-ordinates of the point in the first octant on $l _ { 2 }$ at a distance of $\sqrt { 17 }$ from the point of intersection of $l$ and $l _ { 1 }$ are $( a , b , c )$, then $18 ( a + b + c )$ is equal to $\underline{\hspace{1cm}}$.

Let $P$ be an arbitrary point having sum of the squares of the distance from the planes $x + y + z = 0 , l x - n z = 0$ and $x - 2 y + z = 0$ equal to 9 units. If the locus of the point $P$ is $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 9$, then the value of $l - n$ is equal to

Let $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$ and let $\vec{b}$ be a vector such that $\vec{a} \times \vec{b} = 2\hat{i} - \hat{k}$ and $\vec{a} \cdot \vec{b} = 3$. Then the projection of $\vec{b}$ on the vector $\vec{a} - \vec{b}$ is:
(1) $\frac{2}{\sqrt{21}}$
(2) $2\sqrt{\frac{3}{7}}$
(3) $\frac{2}{3}\sqrt{\frac{7}{3}}$
(4) $\frac{2}{3}$

If $(2, 3, 9)$, $(5, 2, 1)$, $(1, \lambda, 8)$ and $(\lambda, 2, 3)$ are coplanar, then the product of all possible values of $\lambda$ is
(1) $\frac{21}{2}$
(2) $\frac{59}{8}$
(3) $\frac{57}{8}$
(4) $\frac{95}{8}$

Let $\vec { a } = \alpha \hat { i } + \hat { j } + \beta \hat { k }$ and $\vec { b } = 3 \hat { i } - 5 \hat { j } + 4 \hat { k }$ be two vectors, such that $\vec { a } \times \vec { b } = - \hat { i } + 9 \hat { i } + 12 \widehat { k }$. Then the projection of $\vec { b } - 2 \vec { a }$ on $\vec { b } + \vec { a }$ is equal to
(1) 2
(2) $\frac { 39 } { 5 }$
(3) 9
(4) $\frac { 46 } { 5 }$

If the length of the perpendicular drawn from the point $P ( a , 4,2 ) , a > 0$ on the line $\frac { x + 1 } { 2 } = \frac { y - 3 } { 3 } = \frac { z - 1 } { - 1 }$ is $2 \sqrt { 6 }$ units and $Q \left( \alpha _ { 1 } , \alpha _ { 2 } , \alpha _ { 3 } \right)$ is the image of the point $P$ in this line, then $a + \sum _ { i = 1 } ^ { 3 } \alpha _ { i }$ is equal to
(1) 7
(2) 8
(3) 12
(4) 14

Let $S$ be the set of all $a \in R$ for which the angle between the vectors $\vec { u } = a \left( \log _ { e } b \right) \hat { i } - 6 \hat { j } + 3 \hat { k }$ and $\vec { v } = \left( \log _ { e } b \right) \hat { i } + 2 \hat { j } + 2 a \left( \log _ { e } b \right) \hat { k } , ( b > 1 )$ is acute. Then $S$ is equal to
(1) $\left( - \infty , - \frac { 4 } { 3 } \right)$
(2) $\Phi$
(3) $\left( - \frac { 4 } { 3 } , 0 \right)$
(4) $\left( \frac { 12 } { 7 } , \infty \right)$

A plane $E$ is perpendicular to the two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$, and passes through the point $P(1, -1, 1)$. If the distance of the plane $E$ from the point $Q(a, a, 2)$ is $3\sqrt{2}$, then $(PQ)^2$ is equal to
(1) 9
(2) 12
(3) 21
(4) 33

Let $\vec{a}, \vec{b}, \vec{c}$ be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and $(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) + (\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) + (\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b}) = 168$, then $|\vec{a} + \vec{b} + \vec{c}|$ is equal to
(1) 10
(2) 14
(3) 16
(4) 18

If the two lines $l _ { 1 } : \frac { x - 2 } { 3 } = \frac { y + 1 } { - 2 } , z = 2$ and $l _ { 2 } : \frac { x - 1 } { 1 } = \frac { 2 y + 3 } { \alpha } = \frac { z + 5 } { 2 }$ are perpendicular, then an angle between the lines $l _ { 2 }$ and $l _ { 3 } : \frac { 1 - x } { 3 } = \frac { 2 y - 1 } { - 4 } = \frac { z } { 4 }$ is
(1) $\cos ^ { - 1 } \left( \frac { 29 } { 4 } \right)$
(2) $\sec ^ { - 1 } \left( \frac { 29 } { 4 } \right)$
(3) $\cos ^ { - 1 } \left( \frac { 2 } { 29 } \right)$
(4) $\cos ^ { - 1 } \left( \frac { 2 } { \sqrt { 29 } } \right)$

Let the foot of the perpendicular from the point $( 1,2,4 )$ on the line $\frac { x + 2 } { 4 } = \frac { y - 1 } { 2 } = \frac { z + 1 } { 3 }$ be $P$. Then the distance of $P$ from the plane $3 x + 4 y + 12 z + 23 = 0$ is
(1) $\frac { 50 } { 13 }$
(2) $\frac { 63 } { 13 }$
(3) $\frac { 65 } { 13 }$
(4) 4

Let $\vec { a } = 2 \hat { i } - \hat { j } + 5 \hat { k }$ and $\vec { b } = \alpha \hat { i } + \beta \hat { j } + 2 \widehat { k }$. If $( ( \vec { a } \times \vec { b } ) \times \hat { i } ) \cdot \widehat { k } = \frac { 23 } { 2 }$, then $| \vec { b } \times 2 \hat { j } |$ is equal to
(1) 4
(2) 5
(3) $\sqrt { 21 }$
(4) $\sqrt { 17 }$

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) $\frac { a } { 4 } = \frac { b } { 5 } = \frac { c } { 2 }$

The shortest distance between the lines $\frac { x - 3 } { 2 } = \frac { y - 2 } { 3 } = \frac { z - 1 } { - 1 }$ and $\frac { x + 3 } { 2 } = \frac { y - 6 } { 1 } = \frac { z - 5 } { 3 }$ is
(1) $\frac { 18 } { \sqrt { 5 } }$
(2) $\frac { 22 } { 3 \sqrt { 5 } }$
(3) $\frac { 46 } { 3 \sqrt { 5 } }$
(4) $6 \sqrt { 3 }$

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 $\_\_\_\_$.