Answer
( $ t-\displaystyle \frac{1}{2}s+6$ , $s ,\ t$ ),$\ \ s,t\in \mathbb{R}$
Work Step by Step
Write the augmented matrix and,
using row transformations,
arrive at the row-reduced echelon form.
$\left[\begin{array}{llll}
2 & 1 & -2 & 12\\
-1 & -1/2 & 1 & -6\\
3 & 3/2 & -3 & 18
\end{array}\right]\ \ \begin{array}{l}
R_{1}\leftrightarrow R_{2}.\\
.\\
.
\end{array}$
$\left[\begin{array}{llll}
-1 & -1/2 & 1 & -6\\
2 & 1 & -2 & 12\\
3 & 3/2 & -3 & 18
\end{array}\right]\ \ \begin{array}{l}
\times(-1).\\
+2R_{1}.\\
+3R_{1}.
\end{array}$
$\left[\begin{array}{llll}
1 & 1/2 & -1 & 6\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}\right]\ \ \begin{array}{l}
.\\
.\\
.
\end{array}$
Last two rows have all zeros, so
the system is dependent (infinite number of solutions).
Taking
$z=t\in \mathbb{R}$ (arbitrary),
$y=s\in \mathbb{R}$ (arbitrary),
back-substituting:
$x=t-\displaystyle \frac{1}{2}s+6$
Solution set: $\{$( $ t-\displaystyle \frac{1}{2}s+6$ , $s ,\ t$ ),$\ \ s,t\in \mathbb{R} \}$
