JEE Main 2025 — Three Dimensional Geometry Question with Solution

$\begin{array}{rl}& \overrightarrow{\mathrm{p}}=2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}},\overrightarrow{\mathrm{q}}=3\hat{\mathrm{i}}+4\hat{\mathrm{j}}+5\hat{\mathrm{k}}\\ & \Rightarrow \overrightarrow{\mathrm{p}}\times \overrightarrow{\mathrm{q}}=\left|\begin{array}{lll}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|=-\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}}\\ & \mathrm{A}\equiv (1,2,3)B\equiv (\lambda ,4,5)\\ & \text{ Shortest Distance }=\left|\frac{\overrightarrow{\mathrm{AB}}\cdot (\overrightarrow{\mathrm{P}}\times \overrightarrow{\mathrm{q}})}{|\overrightarrow{\mathrm{P}}\times \overrightarrow{\mathrm{q}}|}\right|\end{array}$
$\begin{aligned}
& \frac{1}{\sqrt{6}}=\left|\frac{((\lambda-1) \hat{i}+2 \hat{j}+2 \hat{k}) \cdot(-\hat{i}+2 \hat{j}-\hat{k})}{\sqrt{6}}\right| \\ & \Rightarrow|-\lambda+1+4-2|=1 \Rightarrow|\lambda-3|=1 \\ & \Rightarrow \lambda=3 \pm 1=4,2
\end{aligned}$<br/>Radiusofcirclepassingthroughpoints<br/>$\begin{aligned}
& (0,0),(4,2) \&(2,4) \\ & =\frac{\text { abc }}{4 \Delta}=\frac{\sqrt{20} \times \sqrt{20} \times \sqrt{8}}{4 \times \frac{1}{2}\left|\begin{array}{lll}
1 & 1 & 1 \\ 0 & 4 & 2 \\ 0 & 2 & 4
\end{array}\right|}=\frac{20 \times 2 \sqrt{2}}{2 \times 12} \\ & =\frac{5 \sqrt{2}}{3}
\end{aligned}$