Given: $\mathrm{A}=\left[\begin{array}{lll}2& 0& 1\\ 1& 1& 0\\ 1& 0& 1\end{array}\right]$, $\mathrm{B}=\left[\begin{array}{ccc}{\mathrm{B}}_1& {\mathrm{B}}_2& {\mathrm{B}}_3\end{array}\right]$, ${\mathrm{B}}_1=\left[\begin{array}{l}{\mathrm{x}}_1\\ {\mathrm{y}}_1\\ {\mathrm{z}}_1\end{array}\right]$, ${\mathrm{B}}_2=\left[\begin{array}{l}{\mathrm{x}}_2\\ {\mathrm{y}}_2\\ {\mathrm{z}}_2\end{array}\right]$ and ${\mathrm{B}}_3=\left[\begin{array}{l}{\mathrm{x}}_3\\ {\mathrm{y}}_3\\ {\mathrm{z}}_3\end{array}\right]$.

$\Rightarrow {\mathrm{AB}}_1=\left[\begin{array}{lll}2& 0& 1\\ 1& 1& 0\\ 1& 0& 1\end{array}\right]\left[\begin{array}{l}{\mathrm{x}}_1\\ {\mathrm{y}}_1\\ {\mathrm{z}}_1\end{array}\right]=\left[\begin{array}{l}1\\ 0\\ 0\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}2{\mathrm{x}}_1+0+{\mathrm{z}}_1\\ {\mathrm{x}}_1+{\mathrm{y}}_1+0\\ {\mathrm{x}}_1+0+{\mathrm{z}}_1\end{array}\right]=\left[\begin{array}{l}1\\ 0\\ 0\end{array}\right]$

$\Rightarrow 2{\mathrm{x}}_1+{\mathrm{z}}_1=1, {\mathrm{x}}_1+{\mathrm{y}}_1=0, {\mathrm{x}}_1+{\mathrm{z}}_1=0$

$\Rightarrow {\mathrm{x}}_1=1,{\mathrm{y}}_1=-1,{\mathrm{z}}_1=-1$

$\Rightarrow {\mathrm{AB}}_2=\left[\begin{array}{lll}2& 0& 1\\ 1& 1& 0\\ 1& 0& 1\end{array}\right]\left[\begin{array}{l}{\mathrm{x}}_2\\ {\mathrm{y}}_2\\ {\mathrm{z}}_2\end{array}\right]=\left[\begin{array}{l}2\\ 3\\ 0\end{array}\right]$

$\Rightarrow {\mathrm{AB}}_2=\left[\begin{array}{c}2{\mathrm{x}}_2+{\mathrm{z}}_2\\ {\mathrm{x}}_2+{\mathrm{y}}_2\\ {\mathrm{x}}_2+{\mathrm{z}}_2\end{array}\right]=\left[\begin{array}{l}2\\ 3\\ 0\end{array}\right]$

$\Rightarrow 2{\mathrm{x}}_2+{\mathrm{z}}_2=2, {\mathrm{x}}_2+{\mathrm{y}}_2=3, {\mathrm{x}}_2+{\mathrm{z}}_2=0$

$\Rightarrow {\mathrm{x}}_2=2,{\mathrm{y}}_2=1,{\mathrm{z}}_2=-2$

$\Rightarrow {\mathrm{AB}}_3=\left[\begin{array}{lll}2& 0& 1\\ 1& 1& 0\\ 1& 0& 1\end{array}\right]\left[\begin{array}{l}{\mathrm{x}}_3\\ {\mathrm{y}}_3\\ {\mathrm{z}}_3\end{array}\right]=\left[\begin{array}{l}3\\ 2\\ 1\end{array}\right]$

$\Rightarrow {\mathrm{AB}}_2=\left[\begin{array}{c}2{\mathrm{x}}_3+{\mathrm{z}}_3\\ {\mathrm{x}}_3+{\mathrm{y}}_3\\ {\mathrm{x}}_3+{\mathrm{z}}_3\end{array}\right]=\left[\begin{array}{l}3\\ 2\\ 1\end{array}\right]$

$\Rightarrow 2{\mathrm{x}}_3+{\mathrm{z}}_3=3, {\mathrm{x}}_3+{\mathrm{y}}_3=2, {\mathrm{x}}_3+{\mathrm{z}}_3=1$

$\Rightarrow {\mathrm{x}}_3=2,{\mathrm{y}}_3=0,{\mathrm{z}}_3=-1$

$\Rightarrow \mathrm{B}=\left[\begin{array}{ccc}1& 2& 2\\ -1& 1& 0\\ -1& -2& -1\end{array}\right]$

$\Rightarrow \alpha =\left|\mathrm{B}\right|=3$

$\Rightarrow \beta =1$

$\Rightarrow {\alpha }^3+{\beta }^3=27+1=28$