JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Three Dimensional Geometry

Written in symmetric form the first line is $\frac{x-\lambda }{2}=\frac{y-\frac{1}{2}}{1}=\frac{z}{-1}$, giving the point $A\left(\lambda ,\frac{1}{2},0\right)$ and direction ${\vec{d}}_1=(2,1,-1)$, and the second is $\frac{x}{1}=\frac{y+2\lambda }{1}=\frac{z-\lambda }{1}$, giving $B(0,-2\lambda ,\lambda )$ and ${\vec{d}}_2=(1,1,1)$. Here ${\vec{d}}_1\times {\vec{d}}_2=(2,-3,1)$ with magnitude $\sqrt{14}$, and $\overrightarrow{AB}=\left(-\lambda ,-2\lambda -\frac{1}{2},\lambda \right)$, so $\overrightarrow{AB}\cdot \left({\vec{d}}_1\times {\vec{d}}_2\right)=-2\lambda +6\lambda +\frac{3}{2}+\lambda =5\lambda +\frac{3}{2}$. The shortest distance condition is so $5\lambda =2$ or $5\lambda =-5$. Only the second gives an integer, $\lambda =-1$, so $|\lambda |=1$.