Let $$A = \begin{bmatrix} 1 & 2 \\ 1 & \alpha \end{bmatrix}$$ and $$B = \begin{bmatrix} 3 & 3 \\ \beta & 2 \end{bmatrix}$$. If $$A^2 - 4A + I = O$$ and $$B^2 - 5B - 6I = O$$, then among the two statements : (S1): $$[(B-A)(B+A)]^T = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$$ and (S2): $$\det(\text{adj}(A+B)) = -5$$,

We first determine $$\alpha$$ and $$\beta$$ using the given matrix equations.

For $$A^2 - 4A + I = O$$, we compute $$A^2$$:

$$A^2 = \begin{bmatrix} 1 & 2 \\ 1 & \alpha \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 1 & \alpha \end{bmatrix} = \begin{bmatrix} 3 & 2 + 2\alpha \\ 1 + \alpha & 2 + \alpha^2 \end{bmatrix}$$

So $$A^2 - 4A + I = \begin{bmatrix} 0 & 2\alpha - 6 \\ \alpha - 3 & \alpha^2 - 4\alpha + 3 \end{bmatrix} = O$$

From any entry, $$\alpha = 3$$.

For $$B^2 - 5B - 6I = O$$, we compute $$B^2$$:

$$B^2 = \begin{bmatrix} 3 & 3 \\ \beta & 2 \end{bmatrix}\begin{bmatrix} 3 & 3 \\ \beta & 2 \end{bmatrix} = \begin{bmatrix} 9 + 3\beta & 15 \\ 5\beta & 3\beta + 4 \end{bmatrix}$$

$$B^2 - 5B - 6I = \begin{bmatrix} 3\beta - 12 & 0 \\ 0 & 3\beta - 12 \end{bmatrix} = O$$

So $$\beta = 4$$. Now $$A = \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix}$$ and $$B = \begin{bmatrix} 3 & 3 \\ 4 & 2 \end{bmatrix}$$.

We check statement (S1). We have $$B - A = \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix}$$ and $$B + A = \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix}$$.

$$(B-A)(B+A) = \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix}\begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix} = \begin{bmatrix} 8+5 & 10+5 \\ 12-5 & 15-5 \end{bmatrix} = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$$

Taking the transpose: $$[(B-A)(B+A)]^T = \begin{bmatrix} 13 & 7 \\ 15 & 10 \end{bmatrix}$$

This does not match $$\begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$$, so (S1) is incorrect.

For (S2), we compute $$A + B = \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix}$$ with $$\det(A+B) = 20 - 25 = -5$$.

For a $$2 \times 2$$ matrix, $$\det(\text{adj}(M)) = (\det M)^{n-1} = (-5)^1 = -5$$.

So (S2) is correct.

Hence, the correct answer is Option 2.