Let p and p+2 be prime numbers and let Delta= | vedclass

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(A) Given $\Delta = \left|\begin{array}{ccc} p! & (p+1)! & (p+2)! \ (p+1)! & (p+2)! & (p+3)! \ (p+2)! & (p+3)! & (p+4)!\end{array}ight|$.Taking $p

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$4$

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(A) Given $\Delta = \left|\begin{array}{ccc} p! & (p+1)! & (p+2)! \ (p+1)! & (p+2)! & (p+3)! \ (p+2)! & (p+3)! & (p+4)!\end{array}ight|$.Taking $p!$,$(p+1)!$,and $(p+2)!$ common from rows $1, 2, 3$ respectively:$\Delta = p!(p+1)!(p+2)! \left|\begin{array}{ccc} 1 & p+1 & (p+2)(p+1) \ 1 & p+2 & (p+3)(p+2) \ 1 & p+3 & (p+4)(p+3)\end{array}ight|$.Applying $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_2$:$\Delta = p!(p+1)!(p+2)! \left|\begin{array}{ccc} 1 & p+1 & (p+1)(p+2) \ 0 & 1 & 2(p+2) \ 0 & 1 & 2(p+3)\end{array}ight|$.Expanding along the first column:$\Delta = p!(p+1)!(p+2)! [1 \cdot (2(p+3) - 2(p+2))] = p!(p+1)!(p+2)! [2] = 2 \cdot p!(p+1)!(p+2)!$.Since $p$ and $p+2$ are primes,$p!$ contains $p$ exactly once (for $p \ge 2$).$(p+1)! = (p+1)p!$,so $p$ appears in $p!$ and $(p+1)!$,thus $p^2$ divides $p!(p+1)!$. Also $(p+2)!$ contains $p$ once. So $p^3$ divides $\Delta$,hence $\alpha = 3$.$(p+2)!$ contains $(p+2)$ once. Thus $(p+2)^1$ divides $\Delta$,hence $\beta = 1$.The sum $\alpha + \beta = 3 + 1 = 4$.

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