Finding basis and dimension of a matrix that is not homogeneous? The problem states:
Solve the given system of equations using matrix operations and give the dimension of the solution space.
$x-2y+2z=3$
$-2x+y+2z=1$
The RREF form that I got is:
\begin{bmatrix} 
      1 & 0 & -2 \\ 
      0 & 1 & -2 \\ 
   \end{bmatrix}
The next step is to solve for x and y, which are the basic variables, and I got
$x=2z-5/3$
$y=2z-7/3$
So, my question is, is this the right equation?
$$
\begin{bmatrix} 
x\\
y\\
z
\end{bmatrix}
= s
\begin{bmatrix} 
2\\
2\\
1
\end{bmatrix}
+
\begin{bmatrix} 
-5/3\\
-7/3\\
0
\end{bmatrix}
$$
and the basis?
$$
\begin{bmatrix} 
2\\
2\\
1
\end{bmatrix}
,
\begin{bmatrix} 
-5/3\\
-7/3\\
0
\end{bmatrix}
$$
If not, would it be just 
$$
\begin{bmatrix} 
x\\
y\\
z
\end{bmatrix}
= s
\begin{bmatrix} 
2\\
2\\
1
\end{bmatrix}
$$

Your first option is the correct one. 
We have the equation as
$$\pmatrix{1&-2&2&|&3\\-2&1&2&|&1}$$
You can carry the right hand sides with the row reduction (just consider the row operations acting directly on the equations). We indeed obtain 
$$\pmatrix{1&-2&2&|&3\\0&-3&6&|&7}\ \leadsto\ \pmatrix{1&0&-2&|&-\frac53\\0&1&-2&|&-\frac73}$$
which directly translates back to
$$x-2z=-5/3\\y-2z=-7/3$$
This justifies the first option you gave.
Note, however, that it's not a linear subspace, just an affine one, i.e. it's a translation (by $(0,-\frac53,-\frac73)$)  of a linear subspace (the line $(2,2,1)s$), so its dimension is $1$.
