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Match the following.
LIST-I
LIST-II
A)
Image of the point $(3, 5, 7)$ in the plane
$2 x + y + z = - 18$ is
P)
$(- 1, - 1, - 1)$
B)
The point of intersection of the line $\frac{x - 2}{- 3} = \frac{y - 1}{- 2} = \frac{z - 3}{2}$ and the plane $2 x + y - z = 3$ is
Q)
$(- 21, - 7, - 5)$
C)
The foot of the perpendicular from the point $(1, 1, 2)$ to the plane $2 x - 2 y + 4 z + 5 = 0$ is
R)
$(\frac{23}{10}, \frac{6}{5}, \frac{14}{5})$
D)
The intersection point of the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = z$ is
S)
$(- \frac{1}{12}, \frac{25}{12}, \frac{- 2}{12})$

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$A \to Q; B \to R; C \to S; D \to P$

$A \to P; B \to R; C \to S; D \to Q$

$A \to R; B \to Q; C \to S; D \to P$

$A \to Q; B \to S; C \to R; D \to P$

answer is A.

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Detailed Solution

Image of the point $(3, 5, 7)$ in the plane

$2 x + y + z = - 18$ is

$\frac{x - 3}{2} = \frac{y - 5}{1} = \frac{z - 7}{1} = \frac{- 2 (6 + 5 + 7 + 18)}{4 + 1 + 1}$

$\frac{x - 3}{2} = \frac{y - 5}{1} = \frac{z - 7}{1} = - 12$

$I m a g e = (- 21, - 7, - 5)$

The point of intersection of the line $\frac{x - 2}{- 3} = \frac{y - 1}{- 2} = \frac{z - 3}{2} = t$ and the plane $2 x + y - z = 3$ is

Substituting $(- 3 t + 2, - 2 t + 1, 2 t + 3) i n 2 x + y - z = 3 \Rightarrow t = - \frac{1}{10}$

point of intersection $(\frac{23}{10}, \frac{6}{5}, \frac{14}{5})$

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