JEE Main 2026 Apr 8 Shift 2, Mathematics Q15: Lines in Space

Find $\mu$: $(1,\mu ,2)$ lies on the given line, so $\frac{1-4}{1}=\frac{\mu -9}{2}=\frac{2-5}{1}$. From the first and third, both equal $-3$, giving $\frac{\mu -9}{2}=-3⟹\mu =3$.

Find $\lambda$: The vector from $(\lambda ,2,3)$ to $(1,\mu ,2)=(1,3,2)$ is $(1-\lambda ,1,-1)$ and must be perpendicular to the line's direction $(1,2,1)$:

$(1-\lambda )(1)+1(2)+(-1)(1)=0⟹2-\lambda =0⟹\lambda =2$.

Distance between parallel lines: Both lines have direction $\vec{d}=(2,3,6)$ with $|\vec{d}|=7$. A point on each: $A=(1,2,-4)$, $B=(2,3,-5)$, so $\overrightarrow{AB}=(1,1,-1)$.

$\overrightarrow{AB}\times \vec{d}=\left|\begin{array}{c}\hat{i}\hat{j}\hat{k}\\ 11-1\\ 236\end{array}\right|=\hat{i}(6+3)-\hat{j}(6+2)+\hat{k}(3-2)=(9,-8,1)$.

$|\overrightarrow{AB}\times \vec{d}|=\sqrt{81+64+1}=\sqrt{146}$.

$\therefore$ Distance $=\frac{\sqrt{146}}{7}$.