The system of equations:$\alpha x+y+z=\alpha -1$$x+\alpha y+z=\alpha -1$$x+y+\alpha z=\alpha -1$ has no solution, if $\alpha$ is:

For the system to have no solution or infinite solutions, the determinant of the coefficient matrix must be zero:
$$\Delta =\left|\begin{array}{ccc}\alpha & 1& 1\\ 1& \alpha & 1\\ 1& 1& \alpha \end{array}\right|=0$$
Expanding the determinant:
$$\alpha ({\alpha }^2-1)-1(\alpha -1)+1(1-\alpha )=0$$
$$(\alpha -1)[\alpha (\alpha +1)-1-1]=0$$
$$(\alpha -1)({\alpha }^2+\alpha -2)=0$$
$$(\alpha -1)(\alpha +2)(\alpha -1)=0⟹\alpha =1,-2$$
If $\alpha =1$, the equations become $x+y+z=0$ (all three identical), which has infinitely many solutions.If $\alpha =-2$, the equations become:$-2x+y+z=-3$$x-2y+z=-3$$x+y-2z=-3$ Adding all three equations: $0=-9$, which is a contradiction. Thus, for $\alpha =-2$, there is no solution.