Let $L$ be a line that is perpendicular to the plane $x-3y+2z=k$ and passes through the point $(1,2,-2)$. The point on the line $L$ that is nearest to the $y$ -axis is:

The direction ratios (DRs) of the normal to the plane $x-3y+2z=k$ are $(1,-3,2)$. Since line $L$ is perpendicular to the plane, its DRs are also $(1,-3,2)$.The equation of line $L$ passing through $(1,2,-2)$ is:
$$\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+2}{2}=\lambda$$
Any general point $P$ on line $L$ is $(\lambda +1,-3\lambda +2,2\lambda -2)$.Any point $Q$ on the $y$ -axis is $(0,\mu ,0)$.The DRs of $PQ$ are $(\lambda +1,-3\lambda +2-\mu ,2\lambda -2)$.For the shortest distance, $PQ$ must be perpendicular to line $L$ and the $y$ -axis:Perpendicular to $y$ -axis (DRs $0,1,0$):
$$0(\lambda +1)+1(-3\lambda +2-\mu )+0(2\lambda -2)=0⟹\mu =2-3\lambda$$
Perpendicular to line $L$ (DRs $1,-3,2$):
$$1(\lambda +1)-3(-3\lambda +2-\mu )+2(2\lambda -2)=0$$
$$\lambda +1+9\lambda -6+3\mu +4\lambda -4=0⟹14\lambda +3\mu =9$$
Substitute $\mu =2-3\lambda$:
$$14\lambda +3(2-3\lambda )=9⟹14\lambda +6-9\lambda =9⟹5\lambda =3⟹\lambda =3/5$$
Point $P$ on line $L$ is:
$$\left(\frac{3}{5}+1,-3\left(\frac{3}{5}\right)+2,2\left(\frac{3}{5}\right)-2\right)=\left(\frac{8}{5},\frac{1}{5},-\frac{4}{5}\right)$$