If the system of equations $kx + y + z = 0$, $x + ky - z = 0$, $x - y + z = 0$ has a non-zero solution, then the possible values of $k$ are:

This is a homogeneous system of linear equations (all equations equal zero). For such a system to have a non-zero (non-trivial) solution, the determinant of the coefficient matrix must be equal to zero.

We extract the coefficients of $x, y,$ and $z$ into a $3 \times 3$ matrix:

$\begin{vmatrix} k & 1 & 1 \\ 1 & k & -1 \\ 1 & -1 & 1 \end{vmatrix} = 0$

Expanding along the first row:

$k \begin{vmatrix} k & -1 \\ -1 & 1 \end{vmatrix} - 1 \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & k \\ 1 & -1 \end{vmatrix} = 0$

Now, calculate the $2 \times 2$ determinants:

• $k(k(1) - (-1)(-1)) - 1(1(1) - (1)(-1)) + 1(1(-1) - (1)(k)) = 0$
• $k(k - 1) - 1(1 + 1) + 1(-1 - k) = 0$
• $k^2 - k - 2 - 1 - k = 0$
• $k^2 - 2k - 3 = 0$
• $(k - 3)(k + 1) = 0$

This gives us two possible values:

$k - 3 = 0 \Rightarrow \boxed{k = 3}$

$k + 1 = 0 \Rightarrow \boxed{k = -1}$