For infinite solution | A | = 0 and (adj A) B = 0

⇒ | A | =  $\left|\begin{array}{ccc}1& \mathrm{\alpha }& {\mathrm{\alpha }}^2\\ \mathrm{\alpha }& 1& \mathrm{\alpha }\\ {\mathrm{\alpha }}^2& \mathrm{\alpha }& 1\end{array}\right|=0$

C₁ → C₁ – C₂ and C₂ → C₂ – C₃

$\left|\begin{array}{ccc}-(\mathrm{\alpha }-1)& -\mathrm{\alpha }(\mathrm{\alpha }-1)& {\mathrm{\alpha }}^2\\ (\mathrm{\alpha }-1)& -(\mathrm{\alpha }-1)& \mathrm{\alpha }\\ \mathrm{\alpha }(\mathrm{\alpha }-1)& (\mathrm{\alpha }-1)& 1\end{array}\right|=0$

⇒ (α – 1)2 $\left|\begin{array}{ccc}-1& -\mathrm{\alpha }& {\mathrm{\alpha }}^2\\ 1& -1& \mathrm{\alpha }\\ \mathrm{\alpha }& 1& 1\end{array}\right|=0$

⇒ (α – 1)2 {– 1 (– 1 – α) + α (1 – α2) + α2 (1 + α)} = 0

⇒ (α – 1)2 {α2 + 2α + 1} = 0

⇒ α = 1 or α = – 1

(i)         If α = 1, then plane will be parallel

(ii)        If α = – 1, then A = $\left[\begin{array}{ccc}1& -1& 1\\ -1& 1& -1\\ 1& -1& 1\end{array}\right]$

            adj A = $\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

            $\therefore$ (adj A) B = 0

            $\therefore$ α² + α + 1 = 1.