The correct answer is 117.

Step 1: Identify the given matrices and conditions
We are given the 3 × 3 matrix A:

$A=\left[\begin{array}{ccc}2& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]$
and the matrix B defined by its columns:

$B=\left[\begin{array}{ccc}{B}_1& B_2& B_3\end{array}\right]$
The columns satisfy the following linear equations:

$A·B_1=\left[\begin{array}{c}2\\ 3\\ 1\end{array}\right],A{B}_2=\left[\begin{array}{c}2\\ 0\\ 0\end{array}\right],A{B}_3=\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right]$

Step 2: Find the inverse of matrix A
To determine the column vectors of B, we can compute the inverse matrix of A, denoted as $A^{-1}$.
Let us solve the system $AX=Y$ where:

$\left[\begin{array}{ccc}2& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}{y}_1\\ y_2\\ y_3\end{array}\right]$
This gives the system of equations:
1) $2x_1+x_2=y_1$
2) $x_3=y_2$
3) $x_1=y_3$

Solving for $x_1$, $x_2$, and $x_3$ in terms of $y_1$, $y_2$, and $y_3$:
$x_1=y_3$
$x_2=y_1-2y_3$
$x_3=y_2$

Therefore, the inverse matrix $A^{-1}$ is:

$A^{-1}=\left[\begin{array}{ccc}0& 0& 1\\ 1& 0& -2\\ 0& 1& 0\end{array}\right]$

Step 3: Determine the column vectors of B
Now we calculate $B_1$, $B_2$, and $B_3$:

$B_1=A^{-1}\left[\begin{array}{c}2\\ 3\\ 1\end{array}\right]=\left[\begin{array}{c}0\left(2\right)+0\left(3\right)+1\left(1\right)\\ 1\left(2\right)+0\left(3\right)-2\left(1\right)\\ 0\left(2\right)+1\left(3\right)+0\left(1\right)\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 3\end{array}\right]$

$B_2=A^{-1}\left[\begin{array}{c}2\\ 0\\ 0\end{array}\right]=\left[\begin{array}{c}0\left(2\right)+0\left(0\right)+1\left(0\right)\\ 1\left(2\right)+0\left(0\right)-2\left(0\right)\\ 0\left(2\right)+1\left(0\right)+0\left(0\right)\end{array}\right]=\left[\begin{array}{c}0\\ 2\\ 0\end{array}\right]$

$B_3=A^{-1}\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right]=\left[\begin{array}{c}0\left(3\right)+0\left(2\right)+1\left(1\right)\\ 1\left(3\right)+0\left(2\right)-2\left(1\right)\\ 0\left(3\right)+1\left(2\right)+0\left(1\right)\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ 2\end{array}\right]$

Thus, the matrix B is:

$B=\left[\begin{array}{ccc}1& 0& 1\\ 0& 2& 1\\ 3& 0& 2\end{array}\right]$

Step 4: Compute the determinant and trace of B
1. Determinant of matrix B ($\alpha =\left|B\right|$):
We expand along the first row:
$\left|B\right|=1·\left|\begin{array}{cc}2& 1\\ 0& 2\end{array}\right|-0+1·\left|\begin{array}{cc}0& 2\\ 3& 0\end{array}\right|$
$\left|B\right|=1(4-0)+1(0-6)=4-6=-2$
So, $\alpha =-2$.

2. Diagonal sum (trace) of matrix B ($\beta$):
$\beta =1+2+2=5$

Step 5: Calculate the final value
We need to find the value of ${\alpha }^3+{\beta }^3$:

${\alpha }^3+{\beta }^3={(-2)}^3+5^3$
${\alpha }^3+{\beta }^3=-8+125=117$