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Let Q be the foot of the perpendicular from the point $P (7, - 2, 13)$ on the plane containing the lines $\frac{x + 1}{6} = \frac{y - 1}{7} = \frac{z - 3}{8}$ and $\frac{x - 1}{3} = \frac{y - 2}{5} = \frac{z - 3}{7}$ then ${(P Q)}^{2} =$

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answer is 96.

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Detailed Solution

Eqn.of the plane containing the given lines is

$|\begin{matrix} x + 1 & y - 1 & z - 3 \\ 6 & 7 & 8 \\ 3 & 5 & 7 \end{matrix}| = 0$ $\Rightarrow x - 2 y + z = 0$

$P Q = \frac{|7 - 2 (- 2) + 13|}{\sqrt{1^{2} + 2^{2} + 1^{2}}} = \frac{24}{\sqrt{6}} = 4 \sqrt{6}$

${(P Q)}^{2} = 96$

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