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

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

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

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