Answer
The solution set is $\displaystyle \{(x,y,z) \ |\ x=6-\frac{1}{2}t+u, \ y=t, \ z=u,\ t,u\in \mathbb{R}\}$
(Dependent system)
Work Step by Step
Start with the augmented matrix, row reduce to
reduced row echelon form (Gauss-Jordan.)
$\left[\begin{array}{rrrr}
2 & 1 & -2 & 12\\
-1 & -1/2 & 1 & -6\\
3 & 3/2 & -3 & 18
\end{array}\right]\rightarrow\left(\begin{array}{l}
+R_{2}.\\
.\\
+3R_{2}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{rrrr}
1 & 1/2 & -1 & 6\\
-1 & -1/2 & 1 & -6\\
0 & 0 & 0 & 0
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
+R_{1}.\\
.
\end{array}\right)$
(the last row poses no problem, as it represents 0=0, which is always true)
$\rightarrow\left[\begin{array}{rrrr}
1 & 1/2 & -1 & 6\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}\right]$
The system is consistent (has solutions) and dependent.
$y$ and $z$ have no leading 1 in their corresponding columns.
Let $y=t\in \mathbb{R}, z=u\in \mathbb{R}$, arbitrary numbers,
Row $1 \Rightarrow x=6-\displaystyle \frac{1}{2}t+u$
The solution set is $\displaystyle \{(x,y,z) \ |\ x=6-\frac{1}{2}t+u, \ y=t, \ z=u,\ t,u\in \mathbb{R}\}$
(Dependent system)