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# The system of equations $\alpha x+y+z=\alpha-1$, $x+\alpha y+z=\alpha-1$, $x+y+\alpha z=\alpha-1$ has infinite solutions if $\alpha$ is

$(a)\;-2\qquad(b)\;either \;-2\;or\;1\qquad(c)\;not\;-2\qquad(d)\;1$

$\Delta =\begin{vmatrix}\alpha&1&1\\1&\alpha&1\\1&1&\alpha\end{vmatrix}$
$\;\;\;=\alpha(\alpha^2-1)-1(\alpha-1)+1(1-\alpha)$
$\;\;\;=\alpha(\alpha-1)(\alpha+1)-1(\alpha-1)-1(\alpha-1)$
For infinite solutions $\Delta =0$
$\Rightarrow (\alpha-1)(\alpha^2+\alpha-1-1)=0$
$\Rightarrow (\alpha-1)(\alpha^2+\alpha-2)=0$
$\Rightarrow \alpha=-2,1$
But $\alpha\neq 1$
$\therefore \alpha=-2$
Hence (a) is the correct answer.