Answer
$k=0$, there is 1 solution
$k=3$, there is 0 solution
$k=-3,$ there are $\infty$ solution
Work Step by Step
$kx+3y = 6$
$3x+ky = -6$
They can be converted in the following form
$
\begin{bmatrix}
k && 3 && 6\\
3 && k && -6
\end{bmatrix}
$
This elimination breaks down when k = 0
$\sim
\begin{bmatrix}
0 && 3 && 6\\
3 && 0 && -6
\end{bmatrix}
$
But it can be fixed by a row exchange
$\sim
\begin{bmatrix}
3 && 0 && 6\\
0 && 3 && -6
\end{bmatrix}
$
which on solving gives $3x = 6, 3y = -6$
$x = 2, y = -2$
Hence, one solution
For $ k\neq 0$
$\sim
\begin{bmatrix}
k && 3 && 6\\
3 && k && -6
\end{bmatrix}
$
subtract $3/k$ times the first equation from the second
$\sim
\begin{bmatrix}
k && 3 && 6\\
0 && k-\frac{9}{k} && -6-\frac{18}{k}
\end{bmatrix}
$
$\sim
\begin{bmatrix}
k && 3 && 6\\
0 && \frac{k^2-9}{k} && -6-\frac{18}{k}
\end{bmatrix}
$
The elimination here breaks down if $k = -3,3$
For $k = 3$
$\sim
\begin{bmatrix}
3 && 3 && 6\\
0 && 0 && -12
\end{bmatrix}
$
The second row gives
$0\cdot x+0\cdot y = -12$ which is false
Hence, there are 0 solutions for $k = 3$
For $k = -3$
$\sim
\begin{bmatrix}
-3 && 3 && 6\\
0 && 0 && 0
\end{bmatrix}
$
The first row gives
$-3\cdot x+3\cdot y = 6$ which has infinitely many solutions.
