Let the line passing through the points $P (2, - 1, 2)$ and $Q (5, 3, 4)$ meet the plane $x - y + z = 4$ at the point R. Then the distance of the point from the plane $x + 2 y + 3 z + 2 = 0$ measured parallel to the line $\frac{x - 7}{2} = \frac{y + 3}{2} = \frac{z - 2}{1}$ is equal to

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$\sqrt{61}$

$\sqrt{189}$

$\sqrt{31}$

answer is A.

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Detailed Solution

The equation of PQ is $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2} = t$
Let $R = (3 t + 2, 4 t - 1, 2 t + 2)$ lies on $x - y + z = 4$
$3 t + 2 - 4 t + 1 + 2 t + 2 = 4 \Rightarrow t = - 1 R = (- 1, - 5, 0)$
Equation of line through R, parallel to $\frac{x - 7}{2} = \frac{y + 3}{2} = \frac{z - 2}{1}$ is
$\frac{x + 1}{2} = \frac{y + 5}{2} = \frac{z}{1} = s, Let S = (2 s - 1, 2 s - 5, s)$
$x + 2 y + 3 z + 2 = 0 \Rightarrow 2 s - 1 + 4 s - 10 + 3 s + 2 = 0 \Rightarrow s = 1 \Rightarrow S = (1, - 3, 1)$
$RS = \sqrt{4 + 4 + 1} = 3$