Let Q be the foot of the perpendicular from the point P(7, -2, 13) on the plane containing the lines x + 1 / 6 = y - 1 / 7 = z - 3 / 8 and x - 1 / 3 = y - 2 / 5 = z - 3 / 7. Then (PQ)^(2), is equal to ___________.

Containing the line x + 1 y - 1 z - 3 \ 6 7 8 \ 3 5 7 = 0 9(x + 1) - 18(y - 1) + 9(z - 3) = 0x - 2y + z = 0PQ = | 7 + 4 + 13 / sqrt 6 | = 4sqrt 6 PQ^2 = 96

Let Q be the foot of the perpendicular from the point P... | (Mathematics)

Containing the line $\begin{vmatrix} x + 1 & y - 1 & z - 3 \\ 6 & 7 & 8 \\ 3 & 5 & 7 \end{vmatrix} = 0\\$

$9(x + 1) - 18(y - 1) + 9(z - 3) = 0$

$x - 2y + z = 0$

$PQ = \left| {{{7 + 4 + 13} \over {\sqrt 6 }}} \right| = 4\sqrt 6$

$P{Q^2} = 96$

Explanation

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