In matrix form, the given system is written as
[tex]\begin{bmatrix}1&1&0\\6&2&-3\\k&0&-1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3-k\\1\\-3\end{bmatrix}[/tex]
and the system has no solution is the coefficient matrix is singular. The determinant is
[tex]\begin{vmatrix}1&1&0\\6&2&-3\\k&0&-1\end{vmatrix} = -3k+4[/tex]
and this is zero when k = 4/3.
On the other hand, in case you are missing a factor of z in the first equation, so that the system should read
x + y + (k - 1) z = 2
6x + 2y - 3z = 1
kx - z = -3
reframing it as a matrix equation gives
[tex]\begin{bmatrix}1&1&k-1\\6&2&-3\\k&0&-1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\1\\-3\end{bmatrix}[/tex]
Then the determinant of the coefficient matrix is
[tex]\begin{vmatrix}1&1&k-1\\6&2&-3\\k&0&-1\end{vmatrix} = -2k^2-k+4[/tex]
and the determinant is zero at two values, k = (-1 ± √33)/4.
