QJEEM24Maa9f2: JEE Main 2024 Mathematics - Class 12 JEEM24Maa9f2 Solution: Let the line intersect the lines and be parallel to the line .…

$\begin{aligned}& L_1: \frac{x-2}{1}=\frac{y}{-1}=\frac{z-1}{1}=\lambda \\& L_2: \frac{x+1}{(1 / 2)}=\frac{y-1}{(1 / 2)}=\frac{z+1}{1}=\mu\end{aligned}$

Any point of $L_1$ and $L_2$ will be $(\lambda+2,-\lambda, \lambda+1)$ and $\left(\frac{\mu}{2}-1, \frac{\mu}{2}+1, \mu-1\right)$

Now Dr of line <\lambda-\frac{\mu}{2}+3,-\lambda-\frac{\mu}{2}-1, \lambda-\mu+2>

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

$$\begin{array}{c}\lambda -\frac{\mu }{3}+3=-3\lambda -\frac{3\mu }{2}-3...(1)\\ 2\left(\lambda -\frac{\mu }{3}+3\right)=3(\lambda -\mu +2)...(2)\end{array}\}\lambda =\frac{-4}{3},\mu =\frac{-2}{3}$$

$\therefore \quad$ Points will be $\left(\frac{2}{3}, \frac{4}{3}, \frac{-1}{3}\right)$ and $\left(\frac{-4}{3}, \frac{2}{3}, \frac{-5}{3}\right)$

$\therefore \quad L$ will be $\frac{x-\frac{2}{3}}{3}=\frac{y-\frac{4}{3}}{1}=\frac{z+\frac{1}{3}}{2}$

$\therefore \quad\left(\frac{-1}{3}, 1,-1\right)$ will satisfy $L$