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 $w=[1-(-3),2-0,3-1,2-0,1-2]=(4,2,2,2,-1)$
The norm of $P(w,v)=\frac{}{||v||^2}v$ gives
$P(w,v)=\frac{}{||(0,-2,2,-1,5)||^2}(0,-2,2,-1,5)\\
=\frac{4.0+2.(-2)+2.2+2.(-1)+(-1).5}{0^2+(-2)^2+2^2+(-1)^2+5^2}(0,-2,2,-1,5)\\
=-\frac{7}{34}(0,-2,2,-1,5)\\
=(0,\frac{7}{17},-\frac{7}{17},\frac{7}{34},-\frac{35}{34})$
The distance from $P$ to $L$ is $||w-P(w,v)||\\
=||(4,2,2,2,-1),(0,\frac{7}{17},-\frac{7}{17},\frac{7}{34},-\frac{35}{34})||\\
=||(4,\frac{27}{17},\frac{41}{17},\frac{61}{34},\frac{1}{34})||\\
=\sqrt (4^2+\frac{27}{17})^2+(\frac{41}{17})^2+(\frac{61}{34})^2+(\frac{1}{34})^2=\frac{\sqrt 31858}{34}$
