Let $$A = \begin{matrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{matrix}$$. Then the sum of the diagonal elements of the matrix $$A + I^{11}$$ is equal to:

Given: $$A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{pmatrix}$$. Find the sum of diagonal elements of $$(A + I)^{11}$$.

Adding the identity matrix to $$A$$ gives $$A + I = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 5 & -1 \\ 0 & 12 & -2 \end{pmatrix}.$$

The matrix has a block structure: the (1,1) entry is 2 and the lower-right 2×2 block is $$B = \begin{pmatrix} 5 & -1 \\ 12 & -2 \end{pmatrix}.$$ Hence $$(A+I)^{11} = \begin{pmatrix} 2^{11} & 0 & 0 \\ 0 & & \\ 0 & B^{11} & \end{pmatrix}$$ and its trace is $$2^{11} + \mathrm{tr}(B^{11}).$$

The characteristic polynomial of $$B$$ is $$\det(B - \lambda I) = (5-\lambda)(-2-\lambda) + 12 = \lambda^2 - 3\lambda + 2 = (\lambda-1)(\lambda-2),$$ so the eigenvalues are $$\lambda_1 = 1, \; \lambda_2 = 2.$$

Therefore $$\mathrm{tr}(B^{11}) = \lambda_1^{11} + \lambda_2^{11} = 1 + 2^{11} = 1 + 2048 = 2049.$$

It follows that $$\mathrm{tr}((A+I)^{11}) = 2^{11} + 2049 = 2048 + 2049 = 4097.$$

The correct answer is Option (3): $$\boxed{4097}$$.