Mathematics · 2021
JEE Main · 26 August 2021, Shift 2 · Q89
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…
Let Q be the foot of the perpendicular from the point $\displaystyle \mathrm{P}(7,-2,13)$ on the plane containing the lines $\displaystyle \frac{x+1}{6}=\frac{y-1}{7}=\frac{z-3}{8}$ and $\displaystyle \frac{x-1}{3}=\frac{y-2}{5}=\frac{z-3}{7}$.
Then $\displaystyle (\mathrm{PQ})^2$, is equal to \_\_\_\_.
Official answer
96
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.