JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Three Dimensional Geometry

The normal to the plane containing both lines is $\vec{n}=(\hat{i}+a\hat{j}-\hat{k})\times (-\hat{i}+\hat{j}-a\hat{k})=(1-a^2)\hat{i}+(a+1)\hat{j}+(1+a)\hat{k}=(1-a)(1+a)\hat{i}+(a+1)\hat{j}+(1+a)\hat{k}=(1+a)[(1-a)\hat{i}+\hat{j}+\hat{k}]$. Taking $\vec{n}=(1-a)\hat{i}+\hat{j}+\hat{k}$, and using point $(1,1,0)$ on both lines, the plane equation is $(1-a)(x-1)+(y-1)+z=0$, or $(1-a)x+y+z+a-2=0$. The distance from $(2,1,4)$ is $\frac{|(1-a)\cdot 2+1+4+a-2|}{\sqrt{(1-a{)}^2+1+1}}=\frac{|5-a|}{\sqrt{a^2-2a+3}}=\sqrt{3}$. Squaring: $(5-a{)}^2=3(a^2-2a+3)$, giving $a^2+2a-8=0$, so $a=2$ or $a=-4$. The largest value is $a=2$.