Let the values of $\lambda$ for which the shortest distance between the lines…

Let the values of $\lambda$ for which the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-\lambda}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is $\frac{1}{\sqrt{6}}$ be $\lambda_1$ and $\lambda_2$. Then the radius of the circle passing through the points $(0,0),\left(\lambda_1, \lambda_2\right)$ and $\left(\lambda_2, \lambda_1\right)$ is
  1. $\frac{5 \sqrt{2}}{3}$
  2. 4
  3. $\frac{\sqrt{2}}{3}$
  4. 3

Solution

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

Asked in: JEE Main 2025 (08 Apr Shift 2)

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