Let $$B = \begin{bmatrix} 1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4 \end{bmatrix}$$, $$\alpha > 2$$ be the adjoint of a matrix $$A$$ and $$|A| = 2$$. Then $$\begin{bmatrix} \alpha & -2\alpha & \alpha \end{bmatrix} B \begin{bmatrix} \alpha \\ -2\alpha \end{bmatrix}$$ is equal to

Given $$B = \begin{bmatrix} 1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4 \end{bmatrix}$$, $$\alpha > 2$$, is the adjoint of matrix $$A$$ with $$|A| = 2$$.

Finding $$\alpha$$:

Since $$B = \text{adj}(A)$$, we have $$|B| = |A|^{n-1} = 2^2 = 4$$ (for $$n = 3$$).

Computing $$|B|$$:

$$|B| = 1(8 - 3\alpha) - 3(4 - 3\alpha) + \alpha(\alpha - 2\alpha)$$

$$= 8 - 3\alpha - 12 + 9\alpha - \alpha^2 = -\alpha^2 + 6\alpha - 4$$

Setting $$|B| = 4$$:

$$-\alpha^2 + 6\alpha - 4 = 4 \implies \alpha^2 - 6\alpha + 8 = 0$$

$$(\alpha - 2)(\alpha - 4) = 0 \implies \alpha = 4$$ (since $$\alpha > 2$$)

Computing the expression:

With $$\alpha = 4$$: $$B = \begin{bmatrix} 1 & 3 & 4 \\ 1 & 2 & 3 \\ 4 & 4 & 4 \end{bmatrix}$$, $$\vec{v} = \begin{bmatrix} 4 \\ -8 \\ 4 \end{bmatrix}$$

First compute $$B\vec{v}$$:

$$B\vec{v} = \begin{bmatrix} 4 - 24 + 16 \\ 4 - 16 + 12 \\ 16 - 32 + 16 \end{bmatrix} = \begin{bmatrix} -4 \\ 0 \\ 0 \end{bmatrix}$$

Then $$\vec{v}^T B\vec{v} = [4, -8, 4] \begin{bmatrix} -4 \\ 0 \\ 0 \end{bmatrix} = -16$$

The answer is Option C: $$-16$$.