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Mathematics Question on Matrices

Let p and p + 2 be prime numbers and let
Δ=∣p!(p+1)!(p+2)! (p+1)!(p+2)!(p+3)! (p+2)!(p+3)!(p+4)! ∣Δ=\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 α and β, such that pα and (p + 2)β divide Δ, is _______.

Answer

The correct answer is 4
Δ=∣p!(p+1)!(p+2)! (p+1)!(p+2)!(p+3)! (p+2)!(p+3)!(p+4)! ∣Δ=\begin{vmatrix} p! & (p+1)! & (p+2)! \\\ (p+1)! & (p+2)! & (p+3)! \\\ (p+2)! & (p+3)! & (p+4)! \\\ \end{vmatrix}
=p!(p+1)!⋅(p+2)!$$\begin{vmatrix} 1 & p+1 & (p+1)(p+2) \\\ 1 & (p+2) & (p+2)(p+3) \\\ 1 & (p+3) & (p+3)(p+4) \\\ \end{vmatrix}
=p!(p+1)!⋅(p+2)!$$\begin{vmatrix} 1 & p+1 & p^2+3p+2\\\ 0 & 1 & 2p+4 \\\ 0 & 1 & 2p+6 \\\ \end{vmatrix}
=2(p!)⋅((p+1)!)⋅((p+2)!)=2(p!)⋅((p+1)!)⋅((p+2)!)
=2(p+1)⋅(p!)2⋅((p+2)!)=2(p+1)⋅(p!)2⋅((p+2)!)
=2(p+1)2⋅(p!)3⋅((p+2)!)=2(p+1)2⋅(p!)3⋅((p+2)!)
∴ Maximum value of α is 3 and β is 1.
∴ α + β = 4