JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Three Dimensional Geometry

The given line has direction ratios $2,1,-1$ and passes through $(-1,3,-2)$, so a general point on it is $(-1+2t,3+t,-2-t)$. For the foot of the perpendicular from $P(2,3,1)$ the joining vector $(3-2t,-t,3+t)$ must be perpendicular to the direction, giving $2(3-2t)-t-(3+t)=3-6t=0$, so $t=\frac{1}{2}$ and the foot is $\left(0,\frac{7}{2},-\frac{5}{2}\right)$. The foot is the midpoint of $P$ and its image, so the image is $P^{\prime }(-2,4,-6)$. The required plane contains the second line, which passes through $(2,1,-1)$ with direction $(3,-2,1)$, and also passes through $P^{\prime }$, so it contains the vector $(-4,3,-5)$ joining those two points. A normal to the plane is $(3,-2,1)\times (-4,3,-5)=(7,11,1)$, and using the point $(2,1,-1)$ the plane is $7x+11y+z=24$. Hence $\alpha =7$, $\beta =11$, $\gamma =1$ and $\alpha +\beta +\gamma =19$.