JEE Main 2025 Apr 8 Shift 2, Mathematics Q1: Lines in Space

Take direction vectors $\vec{p}=2\hat{i}+3\hat{j}+4\hat{k}$ and $\vec{q}=3\hat{i}+4\hat{j}+5\hat{k}$; their cross product is $\vec{p}\times \vec{q}=-\hat{i}+2\hat{j}-\hat{k}$, with magnitude $\sqrt{6}$.

Pick $A=(1,2,3)$ on the first line and $B=(\lambda ,4,5)$ on the second, so $\overrightarrow{AB}=(\lambda -1)\hat{i}+2\hat{j}+2\hat{k}$.

Shortest distance formula: $d=\frac{|\overrightarrow{AB}\cdot (\vec{p}\times \vec{q})|}{|\vec{p}\times \vec{q}|}=\frac{|-(\lambda -1)+4-2|}{\sqrt{6}}=\frac{|3-\lambda |}{\sqrt{6}}$.

Setting this equal to $\frac{1}{\sqrt{6}}$ gives $|\lambda -3|=1$, so ${\lambda }_1=4$ and ${\lambda }_2=2$ (or vice versa).

The three points on the circle are $(0,0)$, $(4,2)$ and $(2,4)$. Using the circumradius formula $R=\frac{abc}{4\Delta }$, the side lengths are $\sqrt{20},\sqrt{20},\sqrt{8}$ and the area $\Delta$ computed from the determinant is $6$.

Substituting: $R=\frac{\sqrt{20}\cdot \sqrt{20}\cdot \sqrt{8}}{4\times 6}=\frac{20\times 2\sqrt{2}}{24}=\frac{5\sqrt{2}}{3}$, matching option (1).