Value of k to give matrix infinite, 0, 1 solutions. I have a question that goes: 
For which values of the constant $k$ does the system of equations below have: 


*

*a unique solution,

*no solutions at all,

*infinitely many solutions?


$$
\begin{cases}
x  &- 3y &         &= 6\\
x  &     &+ 3z     &= -3\\
2x &+ ky &+ (3-k)z &= 1
\end{cases}
$$
I tried putting the system of equations into matrix in reduced row echelon form, ended up with the last line being 
$$
\begin{matrix}
0 & k+6 & 3-k &| 1,
\end{matrix}
$$ which I don't think would make sense.

You have
$$
\begin{pmatrix}
1 & -3 & 0   & 6 \\
1 &  0 & 3   & -3 \\
2 &  k & 3-k & 1
\end{pmatrix}
\to
\begin{pmatrix}
1 & -3  & 0   & 6 \\
0 &  3  & 3   & -9 \\
0 & k+6 & 3-k & -11
\end{pmatrix}
\to\\
\begin{pmatrix}
1 &  0  & 3   & -3 \\
0 &  1  & 1   & -3 \\
0 &  0 & 3-k-(k+6) & -11 + 3(k+6)
\end{pmatrix}
\to \\
\begin{pmatrix}
1 &  0  & 3   & -3 \\
0 &  1  & 1   & -3 \\
0 &  0 & -3-2k& 7 + 3k
\end{pmatrix}
$$
What happens in $-3-2k=0$? Can you finish this?
