Let p and p + 2 be prime numbers and let Δ = | cases p ! & (p + 1)...

Given,

\Delta = \left|\begin{cases}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{cases}\right|

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

⇒ \Delta = P ! \left(P + 1\right) ! \left(P + 2\right) ! \left|\begin{cases}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{cases}\right|

Now on solving determinant we get, \left|$\begin{cases}1 & 1 & 1 \\ P + 1 & P + 2 & P + 3 \\ \left(P + 2\right)\left(P + 1\right) & \left(P + 3\right)\left(P + 2\right) & \left(P + 4\right)\left(P + 3\right)\end{cases}$\right|$</p><p>Using operation${C}_{1} \to {C}_{1} - {C}_{2} \text{ and } {C}_{2} \to {C}_{2} - {C}_{3}</p><p>\left|\begin{cases}0 & 0 & 1 \\ - 1 & - 1 & P + 3 \\ - 2P - 4 & - 2P - 6 & \left(P + 4\right)\left(P + 3\right)\end{cases}\right|

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

Now putting the value of determinant we get,

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

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

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

So, $\alpha = 3, \beta = 1$