The system of equations kx + y + z = 1, x + ky + z = k and x + y +...

$kx + y + z = 1$

$x + ky + z = k$

$x + y + zk = {k}^{2}$

\Delta = \left|\begin{cases}K & 1 & 1 \\ 1 & K & 1 \\ 1 & 1 & K\end{cases}\right| = K\left({K}^{2} - 1\right) - 1\left(K - 1\right) + 1\left(1 - K\right)

$= {K}^{3} - K - K + 1 + 1 - K$

$= {K}^{3} - 3K + 2$

$= ( K - 1{)}^{2}\left(K + 2\right)$

For $K = 1$

$\Delta = {\Delta}_{1} = {\Delta}_{2} = {\Delta}_{3} = 0$

But for $K = - 2,$ at least one out of ${\Delta}_{1}, {\Delta}_{2}, {\Delta}_{3}$ are not zero

Hence for no solution, $K = - 2$