Let Q be the foot of the perpendicular from the point P(7,−2,13) on – InterviewSolution

1.	Let Q be the foot of the perpendicular from the point P(7,−2,13) on the plane containing the lines x+16=y−17=z−38 and x−13=y−25=z−37. Then (PQ)2, is equal to
Answer» 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 $(P Q)^{2},$ is equal to

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