Let $$A = \begin{bmatrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$$, $$B = [B_1 \; B_2 \; B_3]$$, where $$B_1, B_2, B_3$$ are column matrices, and $$AB_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$, $$AB_2 = \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}$$, $$AB_3 = \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$$. If $$\alpha = |B|$$ and $$\beta$$ is the sum of all the diagonal elements of $$B$$, then $$\alpha^3 + \beta^3$$ is equal to _______.

We have $$AB = C = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 3 & 2 \\ 0 & 0 & 1 \end{bmatrix}$$, so $$B = A^{-1}C$$.

$$|A| = 2(1) - 0 + 1(0 - 1) = 1$$. $$|C| = 1 \times 3 \times 1 = 3$$ (upper triangular).

$$\alpha = |B| = |A^{-1}||C| = 1 \times 3 = 3$$.

Computing $$A^{-1}$$:

$$A^{-1} = \begin{bmatrix} 1 & 0 & -1 \\ -1 & 1 & 1 \\ -1 & 0 & 2 \end{bmatrix}$$

$$B = A^{-1}C = \begin{bmatrix} 1 & 2 & 2 \\ -1 & 1 & 0 \\ -1 & -2 & -1 \end{bmatrix}$$

$$\beta = \text{tr}(B) = 1 + 1 + (-1) = 1$$.

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