Reply To: Question 18 — JEE Main 21 Jan 2026 S Ii

shahkb4
3 June, 2026 at 8:15 pm #1064

Solution

line $L_{1} \frac{x-2}{-3}=\frac{y-6}{2}=\frac{z-7}{4}=\lambda_{1}$
line $\mathrm{L}_{2} \frac{\mathrm{x}-4}{2}=\frac{\mathrm{y}-3}{1}=\frac{\mathrm{z}-5}{3}=\lambda_{2}$
Point $\mathrm{C}\left(-3 \lambda_{1}+2,2 \lambda_{1}+6,4 \lambda_{1}+7\right)$ on line $\mathrm{L}_{1}$
Point $\mathrm{D}\left(2 \lambda_{2}+4, \lambda_{2}+3,3 \lambda_{2}+5\right)$ on line $\mathrm{L}_{2}$

\[ \begin{aligned} & \mathrm{CD} \xrightarrow{\mathrm{drs}} 2 \lambda_{2}+3 \lambda_{1}+2, \lambda_{2}-2 \lambda_{1}-3,3 \lambda_{2}-4 \lambda_{1}-2 \\ & \quad \text { or }-3,5,16 \\ & \frac{2 \lambda_{2}+3 \lambda_{1}+2}{-3}=\frac{\lambda_{2}-2 \lambda_{1}-3}{5}=\frac{3 \lambda_{2}-4 \lambda_{1}-2}{16} \end{aligned} \]

From first \& middle
$10 \lambda_{2}+15 \lambda_{1}+10=-3 \lambda_{2}+6 \lambda_{1}+9$
$\Rightarrow 13 \lambda_{2}+9 \lambda_{1}+1=0$
From middle \& third
$16 \lambda_{2}-32 \lambda_{1}-48=15 \lambda_{2}-20 \lambda_{1}-10$
$\Rightarrow \lambda_{2}=12 \lambda_{1}+38$
$13\left(12 \lambda_{1}+38\right)+9 \lambda_{1}+1=0$
$\Rightarrow 165 \lambda_{1}+495=0 \Rightarrow \lambda_{1}=-3$
$\Rightarrow \lambda_{2}=-36+38 \Rightarrow \lambda_{2}=2$ Ans.
$\lambda_{1}=-3, \lambda_{2}=2$
we get $\mathrm{C}(11,0,-5) \mathrm{D}(8,5,11)$
$\mathrm{CD}^{2}=9+25+256=290$ Answer(2)