JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Three Dimensional Geometry

The direction of the line of shortest distance is perpendicular to both given lines, so it is the cross product of $(0,1,1)$ and $(2,2,1)$, namely $(-1,2,-2)$, of magnitude $3$. The plane $P:ax-y-z=0$ has normal $(a,-1,-1)$. For a line making an angle $\theta$ with a plane, $\sin \theta$ is the magnitude of the dot product of the line's direction with the plane's unit normal, so $\sin \theta =\frac{|-a-2+2|}{3\sqrt{a^2+2}}=\frac{|a|}{3\sqrt{a^2+2}}$. Equating this to the given $\sqrt{2/27}$ gives $\frac{a^2}{9(a^2+2)}=\frac{2}{27}$, i.e. $3a^2=2(a^2+2)$, so $a^2=4$ and, since $a>0$, $a=2$. The plane is thus $2x-y-z=0$. Reflecting $(1,1,-5)$ in it, evaluate $2(1)-1-(-5)=6$ and step twice the normalised distance along $(2,-1,-1)$: the image is $(1,1,-5)-\frac{2(6)}{6}(2,-1,-1)=(-3,3,-3)$. Hence $\alpha +\beta -\gamma =-3+3-(-3)=3$.