JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Three Dimensional Geometry

From $l+m-n=0$, we get $n=l+m$. Substituting into $l^2+m^2-n^2=0$: $l^2+m^2-(l+m{)}^2=0$, which gives $l^2+m^2-l^2-2lm-m^2=0$, so $-2lm=0$. Thus either $l=0$ or $m=0$. If $l=0$: $m=n$ and direction ratios are $(0,1,1)$. If $m=0$: $l=n$ and direction ratios are $(1,0,1)$. The angle between these lines: $\cos \alpha =\frac{|0\cdot 1+1\cdot 0+1\cdot 1|}{\sqrt{0+1+1}\sqrt{1+0+1}}=\frac{1}{2}$. Therefore $\cos \alpha =\frac{1}{2}$, so ${\cos }^2\alpha =\frac{1}{4}$ and ${\sin }^2\alpha =\frac{3}{4}$. Now ${\sin }^4\alpha +{\cos }^4\alpha =\frac{9}{16}+\frac{1}{16}=\frac{10}{16}=\frac{5}{8}$.