QJEEM24Mab1e8: JEE Main 2024 Mathematics - Class 12 JEEM24Mab1e8 Solution: Let be the point of intersection of the lines and . Then, the…

$\begin{aligned}& L_1: \frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1} \\& L_2: \frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}\end{aligned}$

Point of intersection of $L_1$ and $L_2$ is $(-1,1,-1)$

Distance of point $P$ from $L_3: 4 x=2 y=z$

$L_3: \frac{x}{\frac{1}{4}}=\frac{y}{\frac{1}{2}}=\frac{z}{1}$

Any point on $L_3$ be

$\begin{aligned}& \left(\frac{\lambda}{4}, \frac{\lambda}{2}, \lambda\right) \\& P R:\left\langle\frac{\lambda}{4}+1, \frac{\lambda}{2}-1, \lambda+1\right\rangle \\& \because P R \perp\left\langle\frac{1}{4}, \frac{1}{2}, 1\right\rangle\end{aligned}$

$\begin{aligned}& \Rightarrow\left(\frac{\lambda}{4}+1\right) \frac{1}{4}+\frac{1}{2}\left(\frac{\lambda}{2}-1\right)+\lambda+1=0 \\& \quad \frac{\lambda}{16}+\frac{1}{4}+\frac{\lambda}{4}-\frac{1}{2}+\lambda+1=0 \\& \Rightarrow \quad \lambda=\frac{-4}{7} \\& \therefore \quad R\left(\frac{-1}{7}, \frac{-2}{7}, \frac{-4}{7}\right) \\& \text { Now } R P: \sqrt{\left(\frac{-1}{7}+1\right)^2+\left(\frac{-2}{7}-1\right)^2+\left(\frac{-4}{7}+1\right)^2} \\& \quad=\sqrt{\frac{36}{49}+\frac{81}{49}+\frac{9}{49}}=\frac{\sqrt{126}}{7}=\frac{3 \sqrt{14}}{7}\end{aligned}$