Answer
See below
Work Step by Step
Given $P=(1,2,3,2,1)\\
x(t)=(-3,-2t,1+2t,-4t,2+5t)$
We form vector $w=(9-4,0-6,0-0)=(5,-6,-1)$
The norm of $P(w,v)=\frac{}{||v||^2}v$ gives
$P(w,v)=\frac{}{||(3,0,-1)||^2}(3,0,-1)=(\frac{9}{2},0,-\frac{3}{2})$
The distance from $P$ to $L$ is $||w-P(w,v)||=||w-0||=||w||=||(5,-6,0)-(\frac{9}{2},0,-\frac{3}{2})||=||(\frac{1}{2},-6,\frac{3}{2})||=\sqrt (\frac{1}{2})^2+36+(\frac{3}{2})^2=\sqrt \frac{77}{2}$
