$$\begin{array}{ccc|c}
3&0&-2&-3\\
-2&0&1&-2\\
0&0&-1&2
\end{array}$$
Multiply the first row by 1/3 to put a pivot at $1,1$:
$$\begin{array}{ccc|c}
1&0&-2/3&-1\\
-2&0&1&-2\\
0&0&-1&2
\end{array}$$
Add 2 times first row to the second row to clear out the pivot column:
$$\begin{array}{ccc|c}
1&0&-2/3&-1\\
0&0&-1/3&0\\
0&0&-1&2
\end{array}$$
Multiply the 3rd row by -1 to put a pivot at 3,3:
$$\begin{array}{ccc|c}
1&0&-2/3&-1\\
0&0&-1/3&0\\
0&0&1&-2
\end{array}$$
Clear out third pivot column, first add 2/3 of the 3rd row to the first:
$$\begin{array}{ccc|c}
1&0&0&-7/3\\
0&0&-1/3&0\\
0&0&1&-2
\end{array}$$
Second, add 1/3 of the third row to the second to finish clearing the pivot column:
$$\begin{array}{ccc|c}
1&0&0&-7/3\\
0&0&0&-2/3\\
0&0&1&-2
\end{array}$$
This is the RRE form of your augmented matrix. Note that your equation never had any solutions from the start, as the RRE indicates on the second row: $0 = -2/3$. Also note that most teachers will probably think that adding extra rows and columns of zeros to a matrix is a mistake (and it is if you don't know why it is ok).