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.

Hint: From the first equation we get
$$x=6+3y$$ plug this in the second equation we obtain
$$y+z=-3$$
and with the third equation we obtain
$$(6+k)y+(3-k)z=1$$
with $$y=-3-z$$ we obtain $$(6+k)(-3-z)+(3-k)z=-11$$
so we get
$$z(-k-3)=9+3k$$
Can you finish?
