JEE Main 2025 — 8 April, Evening Shift — Question 29

$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$

$$\begin{gathered} \frac{x-\lambda}{3}=\frac{y-4}{4}=\frac{z-5}{5} \end{gathered}$$ $$\vec{n}_{1} \times \vec{n}_{2}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4\\ 3 & 4 & 5\end{array}\right|$$

$=\hat{i}(15-16)-\hat{j}(10-12)+\hat{k}(8-9)$

$=-\hat{i}+2 \hat{j}-\hat{k}$

$L_{1}$ passing through $(1,2,3)$ and $L_{2}$ through $(\lambda, 4,5)$

$d=\frac{1}{\sqrt{6}}$

$\Rightarrow \frac{|(\lambda-1)(-1)-2(-2)+2(-1)|}{\sqrt{1^{2}+4^{2}+1}}=\frac{1}{\sqrt{6}}$

$|-\lambda+1+4-2|=1$

$|-\lambda+3|=1$

$\lambda-3= \pm 1$

$\lambda=4,2$

Circle passing through $(0,0),(1,4)(4,1)$

$\therefore \quad$ Area $=\frac{1}{2}\left|\begin{array}{lll}0 & 0 & 1\\ 1 & 4 & 1\\ 4 & 1 & 1\end{array}\right|$

$=\left|\frac{1}{2}(1-16)\right|=\frac{15}{2}$

$\therefore r=\frac{a b c}{4 \Delta}$

$=\frac{5 \sqrt{2}}{3}$