JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Three Dimensional Geometry

The rotated plane has equation $(2x+3y+z+20)+\lambda (x-3y+5z-8)=0$. Since it is perpendicular to $2x+3y+z+20=0$, we have $(2+\lambda )(2)+(3-3\lambda )(3)+(1+5\lambda )(1)=0$, giving $\lambda =7$. The rotated plane is $x-2y+4z-4=0$. The line AB through $A\left(2,-\frac{1}{2},2\right)$ perpendicular to this plane is $\frac{x-2}{1}=\frac{y+1/2}{-2}=\frac{z-2}{4}=k$. The midpoint of AB lies on the plane, giving $k=-\frac{2}{3}$. Thus $B\left(\frac{4}{3},\frac{5}{6},-\frac{2}{3}\right)\equiv \left(\frac{8}{6},\frac{5}{6},-\frac{4}{6}\right)$, so $\frac{a}{8}=\frac{b}{5}=\frac{c}{-4}$.