Let the plane $\mathrm{P}: 8 x+\alpha_{1} y+\alpha_{2} z+12=0$ be parallel to

the line $\mathrm{L}: \frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5}$. If the intercept of $\mathrm{P}$

on the $y$-axis is 1 , then the distance between $\mathrm{P}$ and $\mathrm{L}$ is :

P: $8 x+\alpha_{1} \mathrm{y}+\alpha_{2} \mathrm{z}+12=0$ <br/><br/>L: $\frac{\mathrm{x}+2}{2}=\frac{\mathrm{y}-3}{3}=\frac{\mathrm{z}+4}{5}$ <br/><br/>$\because \mathrm{P}$ is parallel to $\mathrm{L}$ <br/><br/>$\Rightarrow 8(2)+\alpha_{1}(3)+5\left(\alpha_{2}\right)=0$ <br/><br/>$\Rightarrow 3 \alpha_{1}+5\left(\alpha_{2}\right)=-16$ <br/><br/>Also $y$-intercept of plane $P$ is 1 <br/><br/>$\Rightarrow \alpha_{1}=-12$ <br/><br/>And $\alpha_{2}=4$ <br/><br/>$\Rightarrow$ Equation of plane $\mathrm{P}$ is $2 \mathrm{x}-3 \mathrm{y}+\mathrm{z}+3=0$ <br/><br/>$\Rightarrow$ Distance of line L from Plane $\mathrm{P}$ is <br/><br/>$=\left|\frac{0-3(6)+1+3}{\sqrt{4+9+1}}\right|$ $=\sqrt{14}$