The distance from the line $L=x-2=\frac{y+3}{2}=\frac{z-1}{4}$ to the plane $P=2y-z=1$, since these lines are parallel, is the distance between any two points from the line and the plane. The line passes through the point $Q=(2,-3,1)$, and the plane has a normal vector $n=2j-k$. The required distance $s$ is $$s=\frac{\vert L \cdot n \vert}{\vert n \vert}=\vert\frac{(0\cdot2)+(-3\cdot-2)+(1\cdot-1)-1}{\sqrt{0^2 +2^2+1^2}}=\frac{8\sqrt{5}}{5}$$
What method can I use to check this answer?
The vector normal of the plane is $\overrightarrow{(0,2,-1)}$ and the point on the line it's $(2,-3,1)$.
Thus, since $\overrightarrow{(1,2,4)}\cdot\overrightarrow{(0,2,-1)}=0$, the distance it's: $$\frac{|0\cdot2+2\cdot(-3)+(-1)\cdot1-1|}{\sqrt{0^2+2^2+(-1)^2}}=\frac{8}{\sqrt5}.$$