For a real number $\alpha ,$ if the system $\left[\begin{array}{ccc}1& \alpha & {\alpha }^2\\ \alpha & 1& \alpha \\ {\alpha }^2& \alpha & 1\end{array}\right] \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right]$ of linear equations, has infinitely many solutions, then $1+\alpha +{\alpha }^2=$

Determinant value of 3$\times$3 matrix should be zero for infinite solution

$\mathrm{\Delta }=0 \Rightarrow 1\left(1-{\alpha }^2\right)-\alpha \left(\alpha -{\alpha }^3\right)+{\alpha }^2\left({\alpha }^2-{\alpha }^2\right)=0$

$\left(1-{\alpha }^2\right)-{\alpha }^2+{\alpha }^4=0$

${\left({\alpha }^2-1\right)}^2=0 \Rightarrow \alpha = \pm 1$

But at  $\alpha =1$, three equations become parallel, so no solution so rejected

at   $\alpha =-1$, All three equation become

$x-y+z=1$  (coincident planes), so infinite solution.

$\therefore 1+\alpha +{\alpha }^2=1$