The first equation in the Reduced Row Echelon Form tells you that we need $x_1+x_2=0$ and the second equation says $x_3=0$. So If we take $x_2=t$ for $t\in\mathbb{R}$ then we must have $x_1=-x_2=-t$ and $x_3=0$.
Thus we have a 1-dimensional solution space determined by the vector
$(x_1,x_2,x_3)=(-1,1,0)$.
This follows from the above discussion since taking $x_2=t$ corresponds to scalar multiplication by t on $(-1,1,0)$.
$$\begin{array}{ccc|c}
3&0&-2&-3\\
-2&0&1&-2\\
0&0&-1&2
\end{array}$$
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).
Best Answer
Sometimes a matrix is just a matrix. How one might interpret an augmented matrix depends a lot on context.
Looking at this from the point of view of this method of computing a matrix inverse, to me it’s more of a bookkeeping device that a representation of any particular system of linear equations. Row-reducing a matrix is equivalent to left-multiplication by some invertible matrix. Because of the way that matrix multiplication is defined this can be “distributed” among portions of the augmented matrix. In particular, you have $$E\left[A\mid I\right] = \left[EA\mid E\right] = [I\mid E].$$ The left-hand side of the result of row-reduction tells us that $EA=I$, or $E=A^{-1}$, which is what is on the right-hand side of the reduced matrix. Augmenting $A$ with the identity matrix is a way of ensuring that the same elementary row operations are automatically applied to both.
Now, one can certainly cook up systems of linear equations that the augmented matrix might represent. As noted in the comments to your question, since the augmentation consists of two columns, it’s tempting to say that it represents two systems of equations, one for each column. However, the context is matrix inversion, which instead suggests to me the matrix equation: $$\begin{bmatrix}1&2\\3&4\end{bmatrix} \begin{bmatrix}m_{11}&m_{12}\\m_{21}&m_{22}\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix},$$ which can be written as the system $$\begin{align} m_{11}+2m_{21} &= 1 \\ 3m_{11}+4m_{21} &= 0 \\ m_{12}+2m_{22} &= 0 \\ 3m_{12}+4m_{22} &= 1.\end{align}$$ This system would more usually be rendered as the augmented matrix $$\left[\begin{array}{cccc|c}1&2&0&0 & 1 \\ 3&4&0&0 & 0 \\ 0&0&1&2 & 0 \\ 0&0&3&4 & 1\end{array}\right]$$ but the repetitive block structure means that you’ll end up applying the same row operations to the last two rows that you do to the first, so this matrix can be collapsed into the more compact form in your question. Note that this system of equations can be broken up into two independent systems of equations, so you could certainly think of it in those terms, too.