Let a line $L$ pass through the point $P(2, 3, 1)$ and be parallel to the line $x + 3y - 2z - 2 = 0 = x - y + 2z$. If the distance of $L$ from the point $(5, 3, 8)$ is $\alpha$, then $3\alpha^{2}$ is equal to $\_\_\_\_$

Let the line $L: \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{1}$ intersect the plane $2x + y + 3z = 16$ at the point $P$. Let the point $Q$ be the foot of perpendicular from the point $R(1,-1,-3)$ on the line $L$. If $\alpha$ is the area of triangle $PQR$, then $\alpha^2$ is equal to $\underline{\hspace{1cm}}$.

Let $\lambda _ { 1 } , \lambda _ { 2 }$ be the values of $\lambda$ for which the points $\left( \frac { 5 } { 2 } , 1 , \lambda \right)$ and $( - 2 , 0 , 1 )$ are at equal distance from the plane $2 x + 3 y - 6 z + 7 = 0$. If $\lambda _ { 1 } > \lambda _ { 2 }$, then the distance of the point $\left( \lambda _ { 1 } - \lambda _ { 2 } , \lambda _ { 2 } , \lambda _ { 1 } \right)$ from the line $\frac { x - 5 } { 1 } = \frac { y - 1 } { 2 } = \frac { z + 7 } { 2 }$ is $\_\_\_\_$

A line with direction ratio $2,1,2$ meets the lines $\mathrm { x } = \mathrm { y } + 2 = \mathrm { z }$ and $\mathrm { x } + 2 = 2 \mathrm { y } = 2 \mathrm { z }$ respectively at the point P and Q . if the length of the perpendicular from the point $( 1,2,12 )$ to the line PQ is $l$, then $l ^ { 2 }$ is

Let the position vectors of the vertices $A , B$ and $C$ of a triangle be $2 \hat { i } + 2 \hat { j } + \hat { k } , \hat { i } + 2 \hat { j } + 2 \hat { k }$ and $2 \hat { i } + \hat { j } + 2 \hat { k }$ respectively. Let $l _ { 1 } , l _ { 2 }$ and $l _ { 3 }$ be the lengths of perpendiculars drawn from the ortho centre of the triangle on the sides $A B , B C$ and $C A$ respectively, then $l _ { 1 } ^ { 2 } + l _ { 2 } ^ { 2 } + l _ { 3 } ^ { 2 }$ equals :
(1) $\frac { 1 } { 5 }$
(2) $\frac { 1 } { 2 }$
(3) $\frac { 1 } { 4 }$
(4) $\frac { 1 } { 3 }$

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 $PQR$ is $\lambda$ and $\lambda ^ { 2 } = 14 K$, then $K$ is equal to:
(1) 36
(2) 81
(3) 72
(4) 18

The square of the distance of the image of the point $( 6,1,5 )$ in the line $\frac { x - 1 } { 3 } = \frac { y } { 2 } = \frac { z - 2 } { 4 }$, from the origin is $\_\_\_\_$

Let a line pass through two distinct points $P ( - 2 , - 1,3 )$ and $Q$, and be parallel to the vector $3 \hat { i } + 2 \hat { j } + 2 \hat { k }$. If the distance of the point Q from the point $\mathrm { R } ( 1,3,3 )$ is 5 , then the square of the area of $\triangle P Q R$ is equal to :
(1) 148
(2) 136
(3) 144
(4) 140

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 in a $\triangle ABC$, the length of the side $AC$ be 6, the vertex $B$ be $(1,2,3)$ and the vertices $A, C$ lie on the line $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$. Then the area (in sq. units) of $\triangle ABC$ is:
(1) 17
(2) 21
(3) 56
(4) 42

Let $P$ be the foot of the perpendicular from the point $Q ( 10 , - 3 , - 1 )$ on the line $\frac { x - 3 } { 7 } = \frac { y - 2 } { - 1 } = \frac { z + 1 } { - 2 }$. Then the area of the right angled triangle $P Q R$, where $R$ is the point $( 3 , - 2,1 )$, is
(1) $9 \sqrt { 15 }$
(2) $\sqrt { 30 }$
(3) $8 \sqrt { 15 }$
(4) $3 \sqrt { 30 }$

If the image of the point $( 4,4,3 )$ in the line $\frac { x - 1 } { 2 } = \frac { y - 2 } { 1 } = \frac { z - 1 } { 3 }$ is $( \alpha , \beta , \gamma )$, then $\alpha + \beta + \gamma$ is equal to
(1) 9
(2) 12
(3) 7
(4) 8

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 foot of the perpendicular from the point $( 1,2,2 )$ on the line $\mathrm { L } : \frac { x - 1 } { 1 } = \frac { y + 1 } { - 1 } = \frac { z - 2 } { 2 }$. Let the line $\vec { r } = ( - \hat { i } + \hat { j } - 2 \hat { k } ) + \lambda ( \hat { i } - \hat { j } + \hat { k } ) , \lambda \in \mathbf { R }$, intersect the line L at Q . Then $2 ( \mathrm { PQ } ) ^ { 2 }$ is equal to :
(1) 25
(2) 19
(3) 29
(4) 27

Q80. Let P be the point of intersection of the lines $\frac { x - 2 } { 1 } = \frac { y - 4 } { 5 } = \frac { z - 2 } { 1 }$ and $\frac { x - 3 } { 2 } = \frac { y - 2 } { 3 } = \frac { z - 3 } { 2 }$. Then, the shortest distance of P from the line $4 x = 2 y = z$ is
(1) $\frac { 5 \sqrt { 14 } } { 7 }$
(2) $\frac { 3 \sqrt { 14 } } { 7 }$
(3) $\frac { \sqrt { 14 } } { 7 }$
(4) $\frac { 6 \sqrt { 14 } } { 7 }$

Q89. Consider a line L passing through the points $\mathrm { P } ( 1,2,1 )$ and $\mathrm { Q } ( 2,1 , - 1 )$. If the mirror image of the point $\mathrm { A } ( 2,2,2 )$ in the line L is $( \alpha , \beta , \gamma )$, then $\alpha + \beta + 6 \gamma$ is equal to $\_\_\_\_$

Q90. Let $P$ be the point $( 10 , - 2 , - 1 )$ and $Q$ be the foot of the perpendicular drawn from the point $R ( 1,7,6 )$ on the line passing through the points $( 2 , - 5,11 )$ and $( - 6,7 , - 5 )$. Then the length of the line segment $P Q$ is equal to

ANSWER KEYS

\begin{tabular}{|l|l|l|l|l|l|l|l|} \hline 1. (1) & 2. (3) & 3. (4) & 4. (2) & 5. (1) & 6. (2) & 7. (3) & 8. (4) \hline 9. (2) & 10. (3) & 11. (2) & 12. (4) & 13. (1) & 14. (3) & 15. (1) & 16. (3) \hline 17. (3) & 18. (1) & 19. (1) & 20. (3) & 21. (4) & 22. (13) & 23. (1) & 24. (12) \hline 25. (2) & 26. (16) & 27. (5) & 28. (250) & 29. (60) & 30. (16) & 31. (4) & 32. (2) \hline 33. (4) & 34. (2) & 35. (3) & 36. (2) & 37. (2) & 38. (2) & 39. (2) & 40. (1) \hline 41. (4) & 42. (3) & 43. (3) & 44. (2) & 45. (4) & 46. (1) & 47. (1) & 48. (4) \hline 49. (1) & 50. (2) & 51. (661) & 52. (5) & 53. (274) & 54. (2) & 55. (76) & 56. (3) \hline 57. (877) & 58. (6) & 59. (8) & 60. (20) & 61. (2) & 62. (4) & 63. (3) & 64. (2) \hline 65. (4) & 66. (1) & 67. (4) & 68. (4) & 69. (3) & 70. (4) & 71. (2) & 72. (4) \hline 73. (1) & 74. (3) & 75. (1) & 76. (2) & 77. (2) & 78. (1) & 79. (2) & 80. (1) \hline

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. 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

Q90. The square of the distance of the image of the point $( 6,1,5 )$ in the line $\frac { x - 1 } { 3 } = \frac { y } { 2 } = \frac { z - 2 } { 4 }$, from the origin is

ANSWER KEYS

\begin{tabular}{|l|l|l|l|} \hline 1. (1) & 2. (2) & 3. (1) & 4. (4) \hline 9. (1) & 10. (1) & 11. (1) & 12. (2) \hline 17. (4) & 18. (1) & 19. (2) & 20. (2) \hline 25. (22) & 26. (16) & 27. (2500) & 28. (28) \hline 33. (1) & 34. (2) & 35. (2) & 36. (4) \hline 41. (4) & 42. (1) & 43. (1) & 44. (3) \hline

In space, a plane E is given with points A and B on it, and a point P at a distance of 4 units from this plane.
The perpendicular projections of line segments PA and PB onto plane E, together with line segment AB, form an equilateral triangle with side length 2 units.
Accordingly, what is the product $| \mathbf { P A } | \cdot | \mathbf { P B } |$?
A) 8 B) 12 C) 16 D) 18 E) 20