JEE Main 2024 — 9 April, Shift 1 — Question 1

$\mathrm{L}_{1}: \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-1}{1}=\lambda$

$\mathrm{L}_{2}: \frac{\mathrm{x}+1}{\frac{1}{2}}=\frac{\mathrm{y}-1}{\frac{1}{2}}=\frac{\mathrm{z}+1}{1}=\mu$

dr of line MN will be $<3+\lambda-\frac{\mu}{2},-1-\lambda-\frac{\mu}{2}, 2+\lambda-\mu>\&$ it will be proportional to $<3,1,2>$

$\therefore \frac{3+\lambda-\frac{\mu}{2}}{3}=\frac{-1-\lambda-\frac{\mu}{2}}{1}=\frac{2+\lambda-\mu}{2}$

$4 \lambda+\mu=-6$ $4+3 \lambda=0$ $\Rightarrow \lambda=-\frac{4}{3} \& \mu=-\frac{2}{3}$

$\therefore$ Coordinate of M will be $<\left(\frac{2}{3}, \frac{4}{3},-\frac{1}{3}\right)$ and equation of required line will be. $\frac{\mathrm{x}-\frac{2}{3}}{3}=\frac{y-\frac{4}{3}}{1}=\frac{z+\frac{1}{3}}{2}=k$

So any point on this line will be $\left(\frac{2}{3}+3 \mathrm{k}, \frac{4}{3}+\mathrm{k},-\frac{1}{3}+2 \mathrm{k}\right)$

$\because \frac{2}{3}+3 \mathrm{k}=-\frac{1}{3} \Rightarrow \mathrm{k}=-\frac{1}{3}$

$\therefore$ Point lie on the line for $\mathrm{k}=-\frac{1}{3}$ is $\left(-\frac{1}{3}, 1,-1\right)$