Given,

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

So, ${}^{n}C_r=28-p, {}^{n}C_{r+1}=p, {}^{n}C_{r+2}=70-\alpha \& {}^{n}C_{r+3}=\alpha$

$\Rightarrow {}^{n}C_r+{}^{n}C_{r+1}=28 \& {}^{n}C_{r+2}+ {}^{n}C_{r+3}=70$

$\Rightarrow {}^{n+1}C_{r+1}=28 \& {}^{n+1}C_{r+3}=70$

We know that, ${}^{8}C_2=28 \text{or} {}^{28}C_1=28$, 

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

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

So, $p={}^{7}C_2=21$ and $\alpha ={}^{7}C_4=35$

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