JEE Main 2026 Jan 24 Shift 2, Mathematics Q11: Lines in Space

Points: $P_1=(-1,2,4)$, $P_2=(0,1,1)$; directions ${\vec{d}}_1=(\alpha ,-1,-\alpha )$, ${\vec{d}}_2=(\alpha ,2,2\alpha )$.

$\overrightarrow{P_1{P}_2}=(1,-1,-3)$.

${\vec{d}}_1\times {\vec{d}}_2=\left|\begin{array}{c}\hat{i}\hat{j}\hat{k}\\ \alpha -1-\alpha \\ \alpha 22\alpha \end{array}\right|=(-2\alpha +2\alpha )\hat{i}-(2{\alpha }^2+{\alpha }^2)\hat{j}+(2\alpha +\alpha )\hat{k}=(0,-3{\alpha }^2,3\alpha )$.

$|{\vec{d}}_1\times {\vec{d}}_2|=3|\alpha |\sqrt{{\alpha }^2+1}$.

Shortest distance: $\frac{|\overrightarrow{P_1{P}_2}\cdot ({\vec{d}}_1\times {\vec{d}}_2)|}{|{\vec{d}}_1\times {\vec{d}}_2|}=\frac{|0+3{\alpha }^2-9\alpha |}{3|\alpha |\sqrt{{\alpha }^2+1}}=\frac{|\alpha -3|}{\sqrt{{\alpha }^2+1}}=\sqrt{2}$.

Square: $(\alpha -3{)}^2=2({\alpha }^2+1)\Rightarrow {\alpha }^2+6\alpha -7=0\Rightarrow (\alpha +7)(\alpha -1)=0$.

$\alpha =-7$ or $\alpha =1$. Sum $=-6$.