Let $\vec{\mathrm{a}} = 3\hat{i} - \hat{j} + 2\hat{k}$, $\vec{\mathrm{b}} = \vec{\mathrm{a}} \times (\hat{i} - 2\hat{k})$ and $\vec{\mathrm{c}} = \vec{\mathrm{b}} \times \hat{k}$. Then the projection of $\vec{\mathrm{c}} - 2\hat{j}$ on $\vec{a}$ is:
(1) $2\sqrt{14}$
(2) $\sqrt{14}$
(3) $3\sqrt{7}$
(4) $2\sqrt{7}$

Let $\mathrm{L}_1: \frac{x-1}{1} = \frac{y-2}{-1} = \frac{z-1}{2}$ and $\mathrm{L}_2: \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z}{1}$ be two lines. Let $L_3$ be a line passing through the point $(\alpha, \beta, \gamma)$ and be perpendicular to both $L_1$ and $L_2$. If $L_3$ intersects $\mathrm{L}_1$, then $|5\alpha - 11\beta - 8\gamma|$ equals:
(1) 20
(2) 18
(3) 25
(4) 16

The perpendicular distance, of the line $\frac { x - 1 } { 2 } = \frac { y + 2 } { - 1 } = \frac { z + 3 } { 2 }$ from the point $\mathrm { P } ( 2 , - 10,1 )$, is :
(1) 6
(2) $5 \sqrt { 2 }$
(3) $4 \sqrt { 3 }$
(4) $3 \sqrt { 5 }$

Let a straight line $L$ pass through the point $P ( 2 , - 1,3 )$ and be perpendicular to the lines $\frac { x - 1 } { 2 } = \frac { y + 1 } { 1 } = \frac { z - 3 } { - 2 }$ and $\frac { x - 3 } { 1 } = \frac { y - 2 } { 3 } = \frac { z + 2 } { 4 }$. If the line $L$ intersects the $y z$-plane at the point $Q$, then the distance between the points $P$ and $Q$ is :
(1) $\sqrt { 10 }$
(2) $2 \sqrt { 3 }$
(3) 2
(4) 3

The square of the distance of the point $\left( \frac { 15 } { 7 } , \frac { 32 } { 7 } , 7 \right)$ from the line $\frac { x + 1 } { 3 } = \frac { y + 3 } { 5 } = \frac { z + 5 } { 7 }$ in the direction of the vector $\hat { i } + 4 \hat { j } + 7 \hat { k }$ is :
(1) 54
(2) 44
(3) 41
(4) 66

Let P be the image of the point $\mathrm{Q}(7, -2, 5)$ in the line $\mathrm{L} : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ and $\mathrm{R}(5, \mathrm{p}, \mathrm{q})$ be a point on $L$. Then the square of the area of $\triangle PQR$ is $\_\_\_\_$.

Q65. If $\mathrm { A } ( 1 , - 1,2 ) , \mathrm { B } ( 5,7 , - 6 ) , \mathrm { C } ( 3,4 , - 10 )$ and $\mathrm { D } ( - 1 , - 4 , - 2 )$ are the vertices of a quadrilateral $A B C D$, then its area is :
(1) $48 \sqrt { 7 }$
(2) $12 \sqrt { 29 }$
(3) $24 \sqrt { 7 }$
(4) $24 \sqrt { 29 }$

Q65. If $A ( 3,1 , - 1 ) , B \left( \frac { 5 } { 3 } , \frac { 7 } { 3 } , \frac { 1 } { 3 } \right) , C ( 2,2,1 )$ and $D \left( \frac { 10 } { 3 } , \frac { 2 } { 3 } , \frac { - 1 } { 3 } \right)$ are the vertices of a quadrilateral $A B C D$, then its area is
(1) $\frac { 2 \sqrt { 2 } } { 3 }$
(2) $\frac { 5 \sqrt { 2 } } { 3 }$
(3) $2 \sqrt { 2 }$
(4) $\frac { 4 \sqrt { 2 } } { 3 }$

Q77. Let $\overrightarrow { \mathrm { a } } = 2 \hat { i } + \hat { j } - \hat { k } , \overrightarrow { \mathrm {~b} } = ( ( \overrightarrow { \mathrm { a } } \times ( \hat { i } + \hat { j } ) ) \times \hat { i } ) \times \hat { i }$. Then the square of the projection of $\overrightarrow { \mathrm { a } }$ on $\overrightarrow { \mathrm { b } }$ is:
(1) $\frac { 1 } { 3 }$
(2) $\frac { 2 } { 3 }$
(3) 2
(4) $\frac { 1 } { 5 }$

Q77. The set of all $\alpha$, for which the vectors $\vec { a } = \alpha t \hat { i } + 6 \hat { j } - 3 \hat { k }$ and $\vec { b } = t \hat { i } - 2 \hat { j } - 2 \alpha t \hat { k }$ are inclined at an obtuse angle for all $t \in \mathbb { R }$, is
(1) $\left( - \frac { 4 } { 3 } , 1 \right)$
(2) $[ 0,1 )$
(3) $\left( - \frac { 4 } { 3 } , 0 \right]$
(4) $( - 2,0 ]$

Q77. Let three vectors $\overrightarrow { \mathrm { a } } = \alpha \hat { i } + 4 \hat { j } + 2 \hat { k } , \overrightarrow { \mathrm {~b} } = 5 \hat { i } + 3 \hat { j } + 4 \hat { k } , \overrightarrow { \mathrm { c } } = x \hat { i } + y \hat { j } + z \hat { k }$ form a triangle such that $\vec { c } = \vec { a } - \vec { b }$ and the area of the triangle is $5 \sqrt { 6 }$. If $\alpha$ is a positive real number, then $| \vec { c } | ^ { 2 }$ is equal to:
(1) 16
(2) 14
(3) 12
(4) 10

Q77. Between the following two statements: Statement I : Let $\vec { a } = \hat { i } + 2 \hat { j } - 3 \hat { k }$ and $\vec { b } = 2 \hat { i } + \hat { j } - \hat { k }$. Then the vector $\vec { r }$ satisfying $\vec { a } \times \vec { r } = \vec { a } \times \vec { b }$ and $\vec { a } \cdot \vec { r } = 0$ is of magnitude $\sqrt { 10 }$. Statement II : In a triangle $A B C , \cos 2 A + \cos 2 B + \cos 2 C \geq - \frac { 3 } { 2 }$.
(1) Statement I is incorrect but Statement II is correct.
(2) Both Statement I and Statement II are correct.
(3) Statement I is correct but Statement II is incorrect.
(4) Both Statement I and Statement II are incorrect.

Q78. Let a unit vector which makes an angle of $60 ^ { \circ }$ with $2 \hat { i } + 2 \hat { j } - \hat { k }$ and angle $45 ^ { \circ }$ with $\hat { i } - \hat { k }$ be $\overrightarrow { \mathrm { C } }$. Then $\overrightarrow { \mathrm { C } } + \left( - \frac { 1 } { 2 } \hat { i } + \frac { 1 } { 3 \sqrt { 2 } } \hat { j } - \frac { \sqrt { 2 } } { 3 } \hat { k } \right)$ is :
(1) $\frac { \sqrt { 2 } } { 3 } \hat { i } - \frac { 1 } { 2 } \hat { k }$
(2) $\left( \frac { 1 } { \sqrt { 3 } } + \frac { 1 } { 2 } \right) \hat { i } + \left( \frac { 1 } { \sqrt { 3 } } - \frac { 1 } { 3 \sqrt { 2 } } \right) \hat { j } + \left( \frac { 1 } { \sqrt { 3 } } + \frac { \sqrt { 2 } } { 3 } \right) \hat { k }$
(3) $\frac { \sqrt { 2 } } { 3 } \hat { i } + \frac { 1 } { 3 \sqrt { 2 } } \hat { j } - \frac { 1 } { 2 } \hat { k }$
(4) $- \frac { \sqrt { 2 } } { 3 } \hat { i } + \frac { \sqrt { 2 } } { 3 } \hat { j } + \left( \frac { 1 } { 2 } + \frac { 2 \sqrt { 2 } } { 3 } \right) \hat { k }$

Q78. If the line $\frac { 2 - x } { 3 } = \frac { 3 y - 2 } { 4 \lambda + 1 } = 4 - z$ makes a right angle with the line $\frac { x + 3 } { 3 \mu } = \frac { 1 - 2 y } { 6 } = \frac { 5 - z } { 7 }$, then $4 \lambda + 9 \mu$ is equal to :
(1) 4
(2) 13
(3) 5
(4) 6

Q78. Let $\overrightarrow { \mathrm { a } } = 6 \hat { i } + \hat { j } - \hat { k }$ and $\overrightarrow { \mathrm { b } } = \hat { i } + \hat { j }$. If $\overrightarrow { \mathrm { c } }$ is a is vector such that $| \overrightarrow { \mathrm { c } } | \geq 6 , \overrightarrow { \mathrm { a } } \cdot \overrightarrow { \mathrm { c } } = 6 | \overrightarrow { \mathrm { c } } | , | \overrightarrow { \mathrm { c } } - \overrightarrow { \mathrm { a } } | = 2 \sqrt { 2 }$ and the angle between $\vec { a } \times \vec { b }$ and $\vec { c }$ is $60 ^ { \circ }$, then $| ( \vec { a } \times \vec { b } ) \times \vec { c } |$ is equal to:
(1) $\frac { 9 } { 2 } ( 6 - \sqrt { 6 } )$
(2) $\frac { 3 } { 2 } \sqrt { 6 }$
(3) $\frac { 9 } { 2 } ( 6 + \sqrt { 6 } )$
(4) $\frac { 3 } { 2 } \sqrt { 3 }$

Q78. Let $\overrightarrow { \mathrm { a } } = \hat { i } + 2 \hat { j } + 3 \hat { k } , \overrightarrow { \mathrm {~b} } = 2 \hat { i } + 3 \hat { j } - 5 \hat { k }$ and $\overrightarrow { \mathrm { c } } = 3 \hat { i } - \hat { j } + \lambda \hat { k }$ be three vectors. Let $\overrightarrow { \mathrm { r } }$ be anit vector along $\vec { b } + \vec { c }$. If $\vec { r } \cdot \vec { a } = 3$, then $3 \lambda$ is equal to:
(1) 21
(2) 30
(3) 25
(4) 27

Q78. Let $\overrightarrow { O A } = 2 \vec { a } , \overrightarrow { O B } = 6 \vec { a } + 5 \vec { b }$ and $\overrightarrow { O C } = 3 \vec { b }$, where $O$ is the origin. If the area of the parallelogram with adjacent sides $\overrightarrow { \mathrm { OA } }$ and $\overrightarrow { \mathrm { OC } }$ is 15 sq. units, then the area (in sq. units) of the quadrilateral OABC is equal to :
(1) 32
(2) 40
(3) 38
(4) 35

Q78. Let $\vec { a } = 2 \hat { i } + \alpha \hat { j } + \hat { k } , \vec { b } = - \hat { i } + \hat { k } , \vec { c } = \beta \hat { j } - \hat { k }$, where $\alpha$ and $\beta$ are integers and $\alpha \beta = - 6$. Let the values of the ordered pair ( $\alpha , \beta$ ), for which the area of the parallelogram of diagonals $\vec { a } + \vec { b }$ and $\vec { b } + \vec { c }$ is $\frac { \sqrt { 21 } } { 2 }$, be ( $\alpha _ { 1 } , \beta _ { 1 }$ ) and $\left( \alpha _ { 2 } , \beta _ { 2 } \right)$. Then $\alpha _ { 1 } ^ { 2 } + \beta _ { 1 } ^ { 2 } - \alpha _ { 2 } \beta _ { 2 }$ is equal to
(1) 19
(2) 17
(3) 24
(4) 21

Q79. Let the point, on the line passing through the points $P ( 1 , - 2,3 )$ and $Q ( 5 , - 4,7 )$, farther from the origin and at distance of 9 units from the point P , be $( \alpha , \beta , \gamma )$. Then $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 }$ is equal to :
(1) 165
(2) 160
(3) 155
(4) 150

Q79. For $\lambda > 0$, let $\theta$ be the angle between the vectors $\vec { a } = \hat { i } + \lambda \hat { j } - 3 \hat { k }$ and $\vec { b } = 3 \hat { i } - \hat { j } + 2 \hat { k }$. If the vectors $\vec { a } + \vec { b }$ and $\vec { a } - \vec { b }$ are mutually perpendicular, then the value of $( 14 \cos \theta ) ^ { 2 }$ is equal to
(1) 50
(2) 40
(3) 25
(4) 20

Q79. Let d be the distance of the point of intersection of the lines $\frac { x + 6 } { 3 } = \frac { y } { 2 } = \frac { z + 1 } { 1 }$ and $\frac { x - 7 } { 4 } = \frac { y - 9 } { 3 } = \frac { z - 4 } { 2 }$ from the point $( 7,8,9 )$. Then $\mathrm { d } ^ { 2 } + 6$ is equal to :
(1) 69
(2) 78
(3) 72
(4) 75

Q79. The shortest distance between the lines $\frac { x - 3 } { 2 } = \frac { y + 15 } { - 7 } = \frac { z - 9 } { 5 }$ and $\frac { x + 1 } { 2 } = \frac { y - 1 } { 1 } = \frac { z - 9 } { - 3 }$ is
(1) $8 \sqrt { 3 }$
(2) $4 \sqrt { 3 }$
(3) $5 \sqrt { 3 }$
(4) $6 \sqrt { 3 }$

Q79. Let $\mathrm { P } ( \alpha , \beta , \gamma )$ be the image of the point $\mathrm { Q } ( 3 , - 3,1 )$ in the line $\frac { x - 0 } { 1 } = \frac { y - 3 } { 1 } = \frac { z - 1 } { - 1 }$ and R be the point $( 2,5 , - 1 )$. If the area of the triangle $P Q R$ is $\lambda$ and $\lambda ^ { 2 } = 14 K$, then $K$ is equal to :
(1) 36
(2) 81
(3) 72
(4) 18

Q79. Let $P ( x , y , z )$ be a point in the first octant, whose projection in the $x y$-plane is the point $Q$. Let $O P = \gamma$; the angle between $O Q$ and the positive $x$-axis be $\theta$; and the angle between $O P$ and the positive $z$-axis be $\phi$, where $O$ is the origin. Then the distance of $P$ from the $x$-axis is
(1) $\gamma \sqrt { 1 - \sin ^ { 2 } \phi \cos ^ { 2 } \theta }$
(2) $\gamma \sqrt { 1 - \sin ^ { 2 } \theta \cos ^ { 2 } \phi }$
(3) $\gamma \sqrt { 1 + \cos ^ { 2 } \phi \sin ^ { 2 } \theta }$
(4) $\gamma \sqrt { 1 + \cos ^ { 2 } \theta \sin ^ { 2 } \phi }$

Q79. Consider the line $L$ passing through the points $( 1,2,3 )$ and $( 2,3,5 )$. The distance of the point $\left( \frac { 11 } { 3 } , \frac { 11 } { 3 } , \frac { 19 } { 3 } \right)$ from the line L along the line $\frac { 3 x - 11 } { 2 } = \frac { 3 y - 11 } { 1 } = \frac { 3 z - 19 } { 2 }$ is equal to
(1) 6
(2) 5
(3) 4
(4) 3