JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Three Dimensional Geometry

The direction vectors are $\vec{p}=(2,3,4)$ and $\vec{q}=(3,4,5)$, with $\vec{p}\times \vec{q}=(-1,2,-1)$ and $|\vec{p}\times \vec{q}|=\sqrt{6}$. Taking $A(1,2,3)$ and $B(\lambda ,4,5)$ on the two lines, $\overrightarrow{AB}=(\lambda -1,2,2)$ and the shortest distance is $\frac{|\overrightarrow{AB}\cdot (\vec{p}\times \vec{q})|}{\sqrt{6}}=\frac{|-(\lambda -1)+4-2|}{\sqrt{6}}=\frac{|3-\lambda |}{\sqrt{6}}$. Equating to $\frac{1}{\sqrt{6}}$ gives $|3-\lambda |=1$, so ${\lambda }_1=4,{\lambda }_2=2$. The circle then passes through $(0,0),(4,2),(2,4)$, whose sides are $2\sqrt{5},2\sqrt{5},2\sqrt{2}$ and whose area is $6$. Hence the circumradius is $R=\frac{(2\sqrt{5})(2\sqrt{5})(2\sqrt{2})}{4\cdot 6}=\frac{40\sqrt{2}}{24}=\frac{5\sqrt{2}}{3}$.