Let $$A = \begin{bmatrix} 2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2 \end{bmatrix}$$ and $$P = \begin{bmatrix} 1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5 \end{bmatrix}$$. The sum of the prime factors of $$|P^{-1}AP - 2I|$$ is equal to

We need to find the sum of prime factors of $$|P^{-1}AP - 2I|$$. For any invertible matrix $$P$$, we use the key property that $$|P^{-1}AP - 2I| = |P^{-1}(A - 2I)P| = |P^{-1}||A - 2I||P| = |A - 2I|$$ because $$P^{-1}AP - 2I = P^{-1}AP - P^{-1}(2I)P = P^{-1}(A - 2I)P$$.

The matrix $$A - 2I$$ equals $$\begin{bmatrix} 2-2 & 1 & 2 \\ 6 & 2-2 & 11 \\ 3 & 3 & 2-2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 6 & 0 & 11 \\ 3 & 3 & 0 \end{bmatrix}$$. To calculate its determinant, we expand along the first row: $$|A - 2I| = 0 \cdot \begin{vmatrix} 0 & 11 \\ 3 & 0 \end{vmatrix} - 1 \cdot \begin{vmatrix} 6 & 11 \\ 3 & 0 \end{vmatrix} + 2 \cdot \begin{vmatrix} 6 & 0 \\ 3 & 3 \end{vmatrix}$$, which simplifies to $$0 - 1(6 \cdot 0 - 11 \cdot 3) + 2(6 \cdot 3 - 0 \cdot 3) = 0 - 1(0 - 33) + 2(18 - 0) = 0 + 33 + 36 = 69$$.

Since $$|P^{-1}AP - 2I| = |A - 2I|$$, it follows that $$|P^{-1}AP - 2I| = 69$$. The prime factorisation of 69 is $$69 = 3 \times 23$$, and thus the sum of its prime factors is $$3 + 23 = 26$$. Therefore, the correct answer is Option 1: 26.