Answer
( $2+3t$ , $t$ , $4$ ),$\ \ t\in \mathbb{R}$
Work Step by Step
Write the augmented matrix and,
using row transformations,
arrive at the row-reduced echelon form.
$\left[\begin{array}{llll}
-2 & 6 & -2 & -12\\
1 & -3 & 2 & 10\\
-1 & 3 & 2 & 6
\end{array}\right] \ \ \begin{array}{l}
\div(-2).\\
-\frac{1}{2}R_{1}.\\
+\frac{1}{2}R_{1}.
\end{array}$
$\left[\begin{array}{llll}
1 & -3 & 1 & 6\\
0 & 0 & 1 & 4\\
0 & 0 & 3 & 12
\end{array}\right] \ \ \begin{array}{l}
-R_{2}.\\
.\\
-3R_{2}.
\end{array}$
$\left[\begin{array}{llll}
1 & -3 & 0 & 2\\
0 & 0 & 1 & 4\\
0 & 0 & 0 & 0
\end{array}\right] \ \ \begin{array}{l}
.\\
.\\
.
\end{array}$
Last row has all zeros, so
the system is dependent (infinite number of solutions).
Taking $y=t\in \mathbb{R}$ (arbitrary),
back-substituting:
$x=2+3t$
$z=4$
Solution set: $\{$( $2+3t$ , $t$ , $4$ ), $t\in \mathbb{R} \}$
