JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Three Dimensional Geometry

The first line passes through $(1,-1,0)$ with direction $(0,-1,1)$, and the second is the intersection of $x+y+z+1=0$ and $2x-y+z+3=0$. A plane through the second line, $x+y+z+1+\lambda (2x-y+z+3)=0$, is parallel to the first line when its normal is perpendicular to $(0,-1,1)$: $(1-\lambda )(-1)+(1+\lambda )(1)=0$ gives $\lambda =0$, so the plane is $x+y+z+1=0$. The shortest distance is then the distance from $(1,-1,0)$ to this plane, $\frac{|1-1+0+1|}{\sqrt{1^2+1^2+1^2}}=\frac{1}{\sqrt{3}}$.