Let p and p + 2 be prime numbers and let

$$ \Delta=\left|\begin{array}{ccc} \mathrm{p} ! & (\mathrm{p}+1) ! & (\mathrm{p}+2) ! \\ (\mathrm{p}+1) ! & (\mathrm{p}+2) ! & (\mathrm{p}+3) ! \\ (\mathrm{p}+2) ! & (\mathrm{p}+3) ! & (\mathrm{p}+4) ! \end{array}\right| $$

Then the sum of the maximum values of $\alpha$ and $\beta$, such that $\mathrm{p}^{\alpha}$ and $(\mathrm{p}+2)^{\beta}$ divide $\Delta$, is __________.

<p>$$\Delta = \left| {\matrix{ {p!} & {(p + 1)!} & {(p + 2)!} \cr {(p + 1)!} & {(p + 2)!} & {(p + 3)!} \cr {(p + 2)!} & {(p + 3)!} & {(p + 4)!} \cr } } \right|$$</p> <p>$$ = p!\,.\,(p + 1)!\,.\,(p + 2)!\left| {\matrix{ 1 & {(p + 1)} & {(p + 1)(p + 2)} \cr 1 & {(p + 2)} & {(p + 2)(p + 3)} \cr 1 & {(p + 3)} & {(p + 3)(p + 4)} \cr } } \right|$$</p> <p>$$ = p!\,.\,(p + 1)!\,.\,(p + 2)!\left| {\matrix{ 1 & {p + 1} & {{p^2} + 3p + 2} \cr 0 & 1 & {2p + 4} \cr 0 & 1 & {2p + 6} \cr } } \right|$$</p> <p>$= 2(p!)\,.\,\left( {(p + 1)!} \right)\,.\,\left( {(p + 2)!} \right)$</p> <p>$= 2(p + 1)\,.\,{(p!)^2}\,.\,\left( {(p + 2)!} \right)$</p> <p>$= 2{(p + 1)^2}\,.\,{(p!)^3}\,.\,\left( {(p + 2)!} \right)$</p> <p>$\therefore$ Maximum value of $\alpha$ is 3 and $\beta$ is 1.</p> <p>$\therefore$ $\alpha + \beta = 4$</p>