Let Q be the foot of perpendicular... | 3D Geometry | JEE Main

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JEE MAIN 2021

26-08-2021 S2

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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 $(P Q)^2$, is equal to $\_\_\_\_$ .

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