Given
$$\begin{bmatrix}
-2 & 4 & -2 & 2 & -2 \\
1 & -2 & 2 & -4 & -1 \\
-1 & 2 & -1 & 1 & -1
\end{bmatrix} \sim \begin{bmatrix}
1 & -2 & 0 & 2 & 3 \\
0 & 0 & 1 & -3 & -2 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix},$$
use the reduced row echelon form above to solve the system
$$\begin{cases}
-2w + 4x - 2y + 2z &= -2 \\
w - 2x + 2y - 4z &= -1 \\
-w + 2x - y + z &= -1
\end{cases}$$
If necessary, parametrize your answer using the free variables of the system.
$$\begin{bmatrix}
w \\
x \\
y \\
z
\end{bmatrix} = $$