If the plane $2x-y+z=0$ is parallel to the line $\frac{2x-1}{2}=\frac{2-y}{2}=\frac{z+1}{a}$, then the value of $a$ is:

Rewrite the line in standard form:
$$\frac{2(x-1/2)}{2}=\frac{-(y-2)}{2}=\frac{z+1}{a}⟹\frac{x-1/2}{1}=\frac{y-2}{-2}=\frac{z+1}{a}$$
The direction vector of the line is $\vec{b}=(1,-2,a)$.The normal vector to the plane $2x-y+z=0$ is $\vec{n}=(2,-1,1)$.If a line is parallel to a plane, the line is perpendicular to the normal of the plane. Thus, $\vec{n}\cdot \vec{b}=0$.
$$(2)(1)+(-1)(-2)+(1)(a)=0$$
$$2+2+a=0⟹a+4=0⟹a=-4$$