Answer
$k=2$ or $k=-1$
Work Step by Step
Set the values to 0, we have:
$x-y+2z=0$
$x+ky+z=0$
$kx+y+3=0$
which is equal to: $\begin{bmatrix}
1 & -1 & 2\\
1 & k & 1\\
k & 1 & 3
\end{bmatrix}$
then we have:
$(-1)^2(-1)\begin{vmatrix}
k & 1\\
1& 3
\end{vmatrix}+(-1)^3(-1)\begin{vmatrix}
1 & 1\\
k & 3
\end{vmatrix}+(-1)^4.2\begin{vmatrix}
1 &k\\
k & 1
\end{vmatrix}$
$0=3k-1+3-k+2-2k^2$
$2(2-k)(1+k)=0$
$2-k=0$ or $1+k=0$
$k=2$ or $k=-1$
