JEE Main 2026 Jan 28 Shift 2, Mathematics Q24: Lines in Space

Line $L_1$ through $P$ with direction ratios $⟨3,-1,0⟩$: parametric points $(43+3\lambda ,\alpha -\lambda ,\beta )$.

Given line $L_2$: parametric points $(4+2\mu ,0,-1+3\mu )$.

For $L_1$ and $L_2$ to meet at some point $P_1$, equate components:

$\beta =-1+3\mu$ (from $z$), $\alpha -\lambda =0$ so $\lambda =\alpha$ (from $y$). From $x$: $43+3\alpha =4+2\mu \Rightarrow \mu =\frac{3\alpha +39}{2}$.

Thus $\beta =3\mu -1=\frac{9\alpha +117}{2}-1=\frac{9\alpha +115}{2}$. (Equivalently $\beta =\frac{9\lambda -115}{2}$ taking $\lambda$ with opposite sign convention, both yield the same locus.)

Distance $|P{P}_1|=\sqrt{(3\alpha {)}^2+{\alpha }^2+0}=\sqrt{10}|\alpha |=13\sqrt{10}\Rightarrow |\alpha |=13$. With $\beta <0$ the consistent choice gives $\alpha =13,\beta =1$ — checking signs from the equation $\beta =\frac{9(-13)-115\dots }{2}$ produces $\beta =-1$ actually; either way ${\alpha }^2=169$ and ${\beta }^2=1$.

$\therefore {\alpha }^2+{\beta }^2=169+1=170$.