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 a system of equations to have no solution, the determinant of coefficients $\Delta$ must be zero.
$$\Delta =\left|\begin{array}{ccc}\alpha & 1& 1\\ 1& \alpha & 1\\ 1& 1& \alpha \end{array}\right|=0$$
Expanding the determinant:
$$\alpha ({\alpha }^2-1)-1(\alpha -1)+1(1-\alpha )=0$$
$$(\alpha -1)[\alpha (\alpha +1)-1-1]=0$$
$$(\alpha -1)[{\alpha }^2+\alpha -2]=0$$
$$(\alpha -1)(\alpha +2)(\alpha -1)=0⟹(\alpha -1{)}^2(\alpha +2)=0$$
So, $\alpha =1$ or $\alpha =-2$.If $\alpha =1$, the equations become $x+y+z=0$ (all three identical), which gives infinitely many solutions.If $\alpha =-2$, checking ${\Delta }_x$:
$${\Delta }_x=\left|\begin{array}{ccc}-3& 1& 1\\ -3& -2& 1\\ -3& 1& -2\end{array}\right|=-3(4-1)-1(6+3)+1(-3-6)=-9-9-9=-27\ne 0$$
Since $\Delta =0$ and ${\Delta }_x\ne 0$ at $\alpha =-2$, the system has no solution.