JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Three Dimensional Geometry

From $2\ell +2m-n=0$, we get $n=2\ell +2m$. Substituting into $mn+n\ell +\ell m=0$: $m(2\ell +2m)+(2\ell +2m)\ell +\ell m=0$ simplifies to $2m\ell +2m^2+2{\ell }^2+2m\ell +\ell m=0$, i.e., $2{\ell }^2+5\ell m+2m^2=0$. Factoring: $(2\ell +m)(\ell +2m)=0$. This gives two lines: (i) $2\ell +m=0$ with $n=2\ell +2m=2\ell -4\ell =-2\ell$, direction ratios $(1,-2,-2)$; (ii) $\ell +2m=0$ with $n=2\ell +2m=-4m+2m=-2m$, direction ratios $(-2,1,-2)$. The dot product $(1)(-2)+(-2)(1)+(-2)(-2)=-2-2+4=0$, so the lines are perpendicular, angle $\pi /2$.