Question:
Let $Q$ be the foot of the perpendicular from the point $\mathrm{P}(7,-2,13)$ on the plane containing the $\operatorname{lines} \frac{x+1}{6}=\frac{y-1}{7}=\frac{z-3}{8}$ and $\frac{x-1}{3}=\frac{y-2}{5}=\frac{z-3}{7}$. Then $(\mathrm{PQ})^{2}$, is equal to___________
Solution:
Containing the line $\left|\begin{array}{ccc}\mathrm{x}+1 & \mathrm{y}-1 & \mathrm{z}-3 \\ 6 & 7 & 8 \\ 3 & 5 & 7\end{array}\right|=0$
$9(x+1)-18(y-1)+9(z-3)=0$
$x-2 y+z=0$
$\mathrm{PQ}=\left|\frac{7+4+13}{\sqrt{6}}\right|=4 \sqrt{6}$
$\mathrm{PQ}^{2}=96$
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