Answer
The solution is, $ w=1,x=-2,y=3\text{ and }z=-4$.
Work Step by Step
Write the augmented matrix, $\left[ \begin{matrix}
3 & 2 & -1 & 2 \\
4 & -1 & 1 & 2 \\
1 & 1 & 1 & 1 \\
-2 & 3 & 2 & -3 \\
\end{matrix}\left| \begin{matrix}
-12 \\
1 \\
-2 \\
10 \\
\end{matrix} \right. \right]$
$\begin{align}
& {{R}_{1}}\leftrightarrow {{R}_{3}} \\
& \left[ \begin{matrix}
1 & 1 & 1 & 1 \\
4 & -1 & 1 & 2 \\
3 & 2 & -1 & 2 \\
-2 & 3 & 2 & -3 \\
\end{matrix}\left| \begin{matrix}
-2 \\
1 \\
-12 \\
10 \\
\end{matrix} \right. \right] \\
\end{align}$
$\begin{align}
& {{R}_{2}}\to {{R}_{2}}-4{{R}_{1}},{{R}_{3}}\to {{R}_{3}}-3{{R}_{1}}+{{R}_{3}}\text{,}{{R}_{4}}\to {{R}_{4}}\text{+}2{{R}_{1}} \\
& \left[ \begin{matrix}
1 & 1 & 1 & 1 \\
0 & -5 & -3 & -2 \\
0 & -1 & -4 & -1 \\
0 & 5 & 4 & -1 \\
\end{matrix}\left| \begin{matrix}
-2 \\
9 \\
-6 \\
6 \\
\end{matrix} \right. \right] \\
\end{align}$
$\begin{align}
& {{R}_{2}}\to -\frac{1}{5}{{R}_{2}} \\
& \left[ \begin{matrix}
1 & 1 & 1 & 1 \\
0 & 1 & \frac{3}{5} & \frac{2}{5} \\
0 & -1 & -4 & -1 \\
0 & 5 & 4 & -1 \\
\end{matrix}\left| \begin{matrix}
-2 \\
-\frac{9}{5} \\
-6 \\
6 \\
\end{matrix} \right. \right] \\
\end{align}$
$\begin{align}
& {{R}_{3}}\to {{R}_{3}}+{{R}_{2}},{{R}_{4}}\to {{R}_{4}}-5{{R}_{2}} \\
& \left[ \begin{matrix}
1 & 1 & 1 & 1 \\
0 & 1 & \frac{3}{5} & \frac{2}{5} \\
0 & 0 & -\frac{17}{5} & -\frac{3}{5} \\
0 & 0 & 1 & -3 \\
\end{matrix}\left| \begin{matrix}
-2 \\
-\frac{9}{5} \\
-\frac{39}{5} \\
15 \\
\end{matrix} \right. \right] \\
\end{align}$
$\begin{align}
& {{R}_{3}}\to -\frac{5}{17}{{R}_{3}} \\
& \left[ \begin{matrix}
1 & 1 & 1 & 1 \\
0 & 1 & \frac{3}{5} & \frac{2}{5} \\
0 & 0 & 1 & \frac{3}{17} \\
0 & 0 & 1 & -3 \\
\end{matrix}\left| \begin{matrix}
-2 \\
-\frac{9}{5} \\
\frac{39}{17} \\
15 \\
\end{matrix} \right. \right] \\
\end{align}$
$\begin{align}
& {{R}_{4}}\to {{R}_{4}}-{{R}_{3}} \\
& \left[ \begin{matrix}
1 & 1 & 1 & 1 \\
0 & 1 & \frac{3}{5} & \frac{2}{5} \\
0 & 0 & 1 & \frac{3}{17} \\
0 & 0 & 0 & -\frac{54}{17} \\
\end{matrix}\left| \begin{matrix}
-2 \\
-\frac{9}{5} \\
\frac{39}{17} \\
\frac{216}{17} \\
\end{matrix} \right. \right] \\
\end{align}$
$\begin{align}
& {{R}_{4}}\to -\frac{17}{54}{{R}_{4}} \\
& \left[ \begin{matrix}
1 & 1 & 1 & 1 \\
0 & 1 & \frac{3}{5} & \frac{2}{5} \\
0 & 0 & 1 & \frac{3}{17} \\
0 & 0 & 0 & 1 \\
\end{matrix}\left| \begin{matrix}
-2 \\
-\frac{9}{5} \\
\frac{39}{17} \\
-4 \\
\end{matrix} \right. \right] \\
\end{align}$
Therefore, the original system is equivalent to, $\begin{align}
& w+x+y+z=-2 \\
& x+\frac{3}{5}y+\frac{2}{5}z=-\frac{9}{5} \\
& y+\frac{3}{17}z=\frac{39}{17} \\
& z=-4 \\
\end{align}$
From the last equation:
$ z=-4$
Substitute zin $ y+\frac{3}{17}z=\frac{39}{17}$:
$\begin{align}
& y-\frac{12}{17}=\frac{39}{17} \\
& y=\frac{51}{17} \\
& y=3 \\
\end{align}$
Substitute z, y in $ x+\frac{3}{5}y+\frac{2}{5}z=-\frac{9}{5}$
$\begin{align}
& x+\frac{9}{5}-\frac{8}{5}=-\frac{9}{5} \\
& x+\frac{1}{5}=-\frac{9}{5} \\
& x=-\frac{10}{5} \\
& x=-2 \\
\end{align}$
Substitute x, y, z in $ w+x+y+z=-2$
$\begin{align}
& w-2+3-4=-2 \\
& w+3=4 \\
& w=1 \\
\end{align}$
The value of variables is:
$\begin{align}
& w=1 \\
& x=-2 \\
& y=3 \\
& z=-4
\end{align}$
Hence, the solution is $ w=1,x=-2,y=3\text{ and }z=-4$.