The system of equations $kx+y+z=1$, $x+ky+z=k$, and $x+y+kz=k^2$ has no solution if $k$ equals:

For the system to have no solution or infinite solutions, the determinant of the coefficient matrix must be zero:
$$\Delta =\left|\begin{array}{ccc}k& 1& 1\\ 1& k& 1\\ 1& 1& k\end{array}\right|=0$$
Expanding the determinant:$k(k^2-1)-1(k-1)+1(1-k)=0$$k^3-k-k+1+1-k=0$$k^3-3k+2=0$ Factoring the cubic equation (by observation, $k=1$ is a root):$(k-1{)}^2(k+2)=0$ So, $k=1$ or $k=-2$.If $k=1$, the equations become:$x+y+z=1$$x+y+z=1$$x+y+z=1$ This results in infinite solutions.If $k=-2$, the equations become inconsistent (no solution).Thus, $k=-2$ is the required value.