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

-2

The correct answer is option (4) : -2

For $\alpha = 1, $ the system reduces to a homogeneous system which is always consistent, So, $\alpha ≠1.$

$D=\begin{vmatrix}\alpha & 1 & 1\\1 & \alpha & 1\\1 & 1 & \alpha\end{vmatrix}=\begin{vmatrix}\alpha +2 & \alpha +2 & \alpha +2\\1 & \alpha & 1\\1 & 1 & \alpha\end{vmatrix}$

[Applying $R_1→R_1+R_2+R_3$]

$⇒D= ( \alpha + 2) \begin{vmatrix}1& 1 & 1\\1 & \alpha & 1\\1 & 1 & \alpha\end{vmatrix}= ( \alpha + 2) \begin{vmatrix}1& 0 & 0\\1 & \alpha -1 & 0\\1 & 0& \alpha -1\end{vmatrix}$

[Applying $C_2→C_2-C_1,C_3→C_3-C_1$]

$⇒D= (\alpha +2) (\alpha -1)^2$

and, $D_1= \begin{vmatrix}\alpha -1& 1 & 1\\\alpha -1 & \alpha & 1\\\alpha -1 & 1 & \alpha\end{vmatrix}=( \alpha -1)\begin{vmatrix}1& 1 & 1\\1 & \alpha & 1\\1 & 1 & \alpha\end{vmatrix}$

$⇒D_1= (\alpha - 1)\begin{vmatrix}1& 0 & 0\\1 & \alpha -1 & 0\\1 & 0 & \alpha -1\end{vmatrix}$   $\begin{bmatrix}Applying \, C_2→C_2-C_1\\C_3→C_3-C_1\end{bmatrix}$

$⇒D_1= (\alpha - 1)^3 $

Clearly, $D=0 $ for $\alpha = -2 $ but $D_1≠0.$

So, the system is inconsistent for $\alpha = -2.$