Mathematics · 2022
JEE Main · 30 June 2022, Shift 1 · Q16
The distance of the point (3,2,-1) from the plane 3 x-y+4 z+1=0 along the line (2-x)/2=(y-3)/2=(z+1)/1 is equal to:
The distance of the point $\displaystyle (3,2,-1)$ from the plane $\displaystyle 3 x-y+4 z+1=0$ along the line $\displaystyle \frac{2-x}{2}=\frac{y-3}{2}=\frac{z+1}{1}$ is equal to :
Official answer
(3)
$\displaystyle 3$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.