JEE Advanced 2021 Paper 1 · Q08 · 3D Geometry

Correct answer:1.5

Solution

Plane $P: x-2y+z-1=0$. Distance from $(0,1,0)$ to $P$ is $\dfrac{|0-2(1)+0-1|}{\sqrt{1^2+(-2)^2+1^2}}=\dfrac{3}{\sqrt{6}}$. Square of distance $D=\dfrac{9}{6}=\dfrac{3}{2}=1.5$.

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