<s>[INST] Write the equations of the system for which $Ax=b$ is compatible $A =\left[ {\begin{array}{cc}
 1&-2&1 \\ -2&1&1 \\ 1&1&-2  \\
\end{array} } \right]
  \ \ \ \
\left[ {\begin{array}{cc}
 1&-2&1&a \\ -2&1&1&b \\ 1&1&-2&c  \\
\end{array} } \right]

  \ \ \ \  
\left[ {\begin{array}{cc}
 1&-2&1&a \\ 0&-3&3&b+2a \\ 0&3&-3&c-a  \\
\end{array} } \right]

  \ \ \ \ 
\left[ {\begin{array}{cc}
 1&-2&1&a \\ 0&-3&3&b+2a \\ 0&0&0&a+b+c  \\
\end{array} } \right]
$
$A$ is the matrix. I concatenated $[A|b]$ and tried to reduce it. I end up with the last matrix, does it mean that the system is not compatible? Is there another way to solve it?
Or should I concatenate $[A|0]$ to make a homogeneous system?    
UPDATE: 
$A =\left[ {\begin{array}{cc}
 1&-2&1 \\ -2&1&1 \\ 1&1&-2  \\
\end{array} } \right]
  \ \ \ \
\left[ {\begin{array}{cc}
 1&-2&1&0 \\ -2&1&1&0 \\ 1&1&-2&0  \\
\end{array} } \right]
  \ \ \ \ 
\left[ {\begin{array}{cc}
 1&-2&1&0 \\ 0&-3&3&0 \\ 0&0&0&0  \\
\end{array} } \right]
\ \ \ 
\left[ {\begin{array}{cc}
 1&0&-1&0 \\ 0&1&-1&0 \\ 0&0&0&0  \\
\end{array} } \right]
$
After solving $[A|0]$ I get $a-c=0$ and $b-c=0$, the equations of the system seem to be: $a=c,b=c$, so $a=b=c$. Is this the correct solution?
 [/INST] What you did in the Update just shows that  $A{\bf x}=\bf0$ has one (independent) solution. That is the dimension of the null space of $A$ is 1. 
So $A{\bf x}=\bf b$ has two  independent solutions. That is, the dimension of the column space of $A$ is 2. You have an echelon form of $A$. What you need to do is find a basis for the column space of $A$. Then the set of linear combinations of elements of this basis will be the answer to your problem (that is $\bf y$ is in the column space of $A$ if and only if $A{\bf x}=\bf y$ has a solution).
That is, if I'm interpreting the problem correctly: for what vectors $\bf b$ does $A{\bf x}=\bf b$ have a solution?
You could also keep going with what you did at the start: find conditions on $a$, $b$, and $c$ so that the system represented by $\left[ {\begin{array}{cc}
 1&-2&1&a \\ 0&-3&3&b+2a \\ 0&0&0&a+b+c  \\
\end{array} } \right]$ has a solution.
</s>