JEE Main 2026 Apr 4 Shift 1, Mathematics Q14: Lines in Space

Parametrise: $P=(2\lambda ,3\lambda ,3\lambda -1)$ on the first line; $Q=(-\mu -1,\mu +2,4\mu )$ on the second.

The vector $\overrightarrow{PQ}$ must be parallel to $(1,-1,2)$, so

$\frac{-\mu -1-2\lambda }{1}=\frac{\mu +2-3\lambda }{-1}=\frac{4\mu -3\lambda +1}{2}.$

From the first two: $-(-\mu -1-2\lambda )=\mu +2-3\lambda \Rightarrow \mu +1+2\lambda =\mu +2-3\lambda \Rightarrow 5\lambda =1$, so $\lambda =\frac{1}{5}$.

From the first and third: $2(-\mu -1-2\lambda )=4\mu -3\lambda +1\Rightarrow -2\mu -2-4\lambda =4\mu -3\lambda +1\Rightarrow -6\mu =3+\lambda =\frac{16}{5}$, so $\mu =-\frac{8}{15}$.

$P=\left(\frac{2}{5},\frac{3}{5},-\frac{2}{5}\right),Q=\left(-\frac{7}{15},\frac{22}{15},-\frac{32}{15}\right).$

$\overrightarrow{PQ}=\left(-\frac{13}{15},\frac{13}{15},-\frac{26}{15}\right)$, so

${\alpha }^2=\frac{169+169+676}{225}=\frac{1014}{225}.$

$\therefore 225{\alpha }^2=1014.$