For a real number α , if the system 1 α α 2 α 1 α α 2 α 1   x y z =…

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.

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