$\text{ Find the distance of the point }\mathrm{P}(3,8,2)\text{ from the line }\frac{1}{2}(\mathrm{x}-1)=\frac{1}{4}(\mathrm{y}-3)=\frac{1}{3}(\mathrm{z}-2)\text{ measured }$

$\text{ parallel to the plane }3x+2y-2z+15=0$

$\text{ Let general point of line be }A(2\lambda +1,4\lambda +3,3\lambda +2)\text{. Let this point lies at the same distance as }$

$\text{ the point p }(3,8,2)\text{ from the plane }3x+2y-2z+15=0$

$\text{ Therefore, }\frac{3.3+2.8-2.2+15}{\sqrt{17}}=\frac{3(2\lambda +1)+2(4\lambda +3)-2(3\lambda +2)+15}{\sqrt{17}}$

$\begin{array}{l}\Rightarrow 36=8\lambda +20\Rightarrow \lambda =2\\ \text{ Therefore, }A\text{ is }(5,11,8)\\ PA=\sqrt{(5-3{)}^2+(11-8{)}^2+(8-2{)}^2}=\sqrt{4+9+36}=7\end{array}$