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…

Given,

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

Now using the factorial concept and taking common terms we get,

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

Now on solving determinant we get,  $\left|\begin{array}{ccc}1& 1& 1\\ P+1& P+2& P+3\\ \left(P+2\right)\left(P+1\right)& \left(P+3\right)\left(P+2\right)& \left(P+4\right)\left(P+3\right)\end{array}\right|$

Using operation $C_1\to C_1-C_2 \& C_2\to C_2-C_3$

$\left|\begin{array}{ccc}0& 0& 1\\ -1& -1& P+3\\ -2P-4& -2P-6& \left(P+4\right)\left(P+3\right)\end{array}\right|$

$=2P+6-\left(2P+4\right)=2$

Now putting the value of determinant we get,

$\Rightarrow \Delta =2P!\left(P+1\right)!\left(P+2\right)!$

$\Rightarrow \Delta =2P^3\left(P-1\right)!\left(P+1\right)\left(P-1\right)!\left(P+2\right)\left(P+1\right)\left(P-1\right)!$

Which is divisible by $P^{\alpha }$ and ${\left(P+2\right)}^{\beta }$

So, $\alpha =3,\beta =1$