Answer
The solution set is $\{(2+3t\ ,\ t,\ 4) \ |\ \ t\in \mathbb{R}\}$
(Dependent system)
Work Step by Step
Start with the augmented matrix, row reduce to
reduced row echelon form (Gauss-Jordan.)
$\left[\begin{array}{llll}
-2 & 6 & -2 & -12\\
1 & -3 & 2 & 10\\
-1 & 3 & 2 & 6
\end{array}\right]\rightarrow\left(\begin{array}{l}
\leftrightarrow R_{2}.\\
.\\
+R_{2}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & -3 & 2 & 10\\
-2 & 6 & -2 & -12\\
0 & 0 & 4 & 16
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
+2R_{1}.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & -3 & 2 & 10\\
0 & 0 & 2 & 8\\
0 & 0 & 4 & 16
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
\div 2.\\
-2R_{2}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{llll}
1 & -3 & 2 & 10\\
0 & 0 & 1 & 4\\
0 & 0 & 0 & 0
\end{array}\right]$
The last row represents 0=0, which is always true.
The system is consistent (has solutions) and dependent.
$y$ has no leading 1 in its corresponding column.
Let $y=t\in \mathbb{R}$, an arbitrary number,
Row $2 \Rightarrow z=4$
Row $1 \Rightarrow x-3t+2(4)=10 \Rightarrow x=2+3t$
The solution set is $\{(2+3t\ ,\ t,\ 4) \ |\ \ t\in \mathbb{R}\}$
(Dependent system)