Suppose 28 - p, p, 70 - α, α are the coefficient of four...

Given,

$28 - p, p, 70 - \alpha, \alpha$ are the coefficient of four consecutive terms in the expansion of $( 1 + x{)}^{n}$

So, $Cr n = 28 - p, Cr + 1 n = p, Cr + 2 n = 70 - \alpha \text{\text{ and } } Cr + 3 n = \alpha$

$⇒ Cr n + Cr + 1 n = 28 \text{\text{ and } }$ Cr + 2 n + Cr + 3 n = 70

$⇒ Cr + 1 n + 1 = 28 \text{\text{ and } }$ Cr + 3 n + 1 = 70

We know that, $C2 8 = 28\text{or} C1 28 = 28$, 

Now on comparing with $Cr + 1 n + 1 = 28$ we get, $\left(n, r\right) =$\left(7, 1\right)\text{or}\left(n, r\right)$= \left(27, 0\right)$

Now, checking with $Cr + 3 n + 1 = 70$ we get, \left(n, r\right)$= \left(7, 1\right)$

So, p = C2 7 = 21 and $\alpha = C4 7 = 35$

 Hence, the value of $2\alpha - 3p = 70 - 63 = 7$