Let p and p+2 be prime numbers and let $$\Delta = \begin{vmatrix} p! & (p+1)! & (p+2)! \\ (p+1)! & (p+2)! & (p+3)! \\ (p+2)! & (p+3)! & (p+4)! \end{vmatrix}$$
Then the sum of the maximum values of $$\alpha$$ and $$\beta$$, such that $$p^\alpha$$ and $$(p+2)^\beta$$ divide $$\Delta$$, is _______

We are given that $$p$$ and $$p+2$$ are both prime (a twin prime pair), and the determinant $$\Delta = \begin{vmatrix} p! & (p+1)! & (p+2)! \\ (p+1)! & (p+2)! & (p+3)! \\ (p+2)! & (p+3)! & (p+4)! \end{vmatrix}$$. We need the sum of the maximum values of $$\alpha$$ and $$\beta$$ such that $$p^\alpha$$ and $$(p+2)^\beta$$ divide $$\Delta$$.

We factor out $$p!$$ from Row 1, $$(p+1)!$$ from Row 2, and $$(p+2)!$$ from Row 3 to obtain:

$$\Delta = p! \cdot (p+1)! \cdot (p+2)! \begin{vmatrix} 1 & p+1 & (p+1)(p+2) \\ 1 & p+2 & (p+2)(p+3) \\ 1 & p+3 & (p+3)(p+4) \end{vmatrix}$$

We simplify the remaining determinant by performing $$R_2 \to R_2 - R_1$$ and $$R_3 \to R_3 - R_1$$:

$$\begin{vmatrix} 1 & p+1 & (p+1)(p+2) \\ 0 & 1 & 2(p+2) \\ 0 & 2 & 2(2p+5) \end{vmatrix}$$

where we used $$(p+2)(p+3) - (p+1)(p+2) = (p+2) \cdot 2 = 2(p+2)$$ and $$(p+3)(p+4) - (p+1)(p+2) = 4p + 10 = 2(2p+5)$$.

Expanding along column 1, the determinant equals $$1 \cdot [1 \cdot 2(2p+5) - 2 \cdot 2(p+2)] = 2(2p+5) - 4(p+2) = 4p + 10 - 4p - 8 = 2$$.

Therefore $$\Delta = 2 \cdot p! \cdot (p+1)! \cdot (p+2)!$$.

Now we determine the maximum power of $$p$$ dividing $$\Delta$$. Since $$p$$ is prime, $$p!$$ contains $$p$$ as a factor exactly once, and $$(p+1)! = (p+1) \cdot p!$$ also contains exactly one factor of $$p$$ (as $$p+1$$ is not divisible by $$p$$). Similarly $$(p+2)! = (p+2)(p+1) \cdot p!$$ contains exactly one factor of $$p$$ (since neither $$p+1$$ nor $$p+2$$ is divisible by prime $$p \geq 3$$). The factor 2 does not contribute any power of $$p$$. So the maximum $$\alpha = 1 + 1 + 1 = 3$$.

For the maximum power of $$(p+2)$$ dividing $$\Delta$$: since $$p + 2$$ is prime and $$p + 2 > p$$, neither $$p!$$ nor $$(p+1)!$$ contains the factor $$p+2$$. However $$(p+2)!$$ contains exactly one factor of $$p+2$$. So the maximum $$\beta = 1$$.

The sum of the maximum values is $$\alpha + \beta = 3 + 1 = 4$$.

Hence, the correct answer is 4.