JEE Main 2023 (Online) 15th April Morning ShiftQuestion 1
Let the foot of perpendicular of the point $P(3,-2,-9)$ on the plane passing through the points $(-1,-2,-3)$, $(9,3,4)$, $(9,-2,1)$ be $Q(\alpha,\beta,\gamma)$. Then the distance of $Q$ from the origin is
3D Geometry
Solution
Vectors $(10,5,7)$ and $(10,0,4)$ from $(-1,-2,-3)$. Normal $=(5\cdot4-7\cdot0,\ 7\cdot10-10\cdot4,\ 0-50)=(20,30,-50)\parallel(2,3,-5)$. Plane: $2(x+1)+3(y+2)-5(z+3)=0\Rightarrow 2x+3y-5z=7$. For $P(3,-2,-9)$: $6-6+45-7=38$, $|n|^2=38$, so $Q=P-1\cdot(2,3,-5)=(1,-5,-4)$. $OQ^2=1+25+16=42$. Option (D).