Given a plane $P$ is parallel to two lines whose direction ratios are $-2,1,-3$, and $-1,2,-2$,

So normal of plane $P$ is given by, $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& 1& -3\\ -1& 2& -2\end{array}\right|=4\hat{i}-\hat{j}-3\hat{k}$

Now equation of plane $P$ which passes through $\left(2,2,-2\right)$ will be,

$4\left(\mathrm{x}-2\right)-\left(\mathrm{y}-2\right)-3\left(\mathrm{z}+2\right)=0$

$\Rightarrow 4\mathrm{x}-\mathrm{y}-3\mathrm{z}-12=0$

Now, coordinates of plane making intercepts at $X,Y \& Z$ axis will be  $A\left(3,0,0\right),B\left(0,-12,0\right),C\left(0,0,-4\right)$

$\Rightarrow \alpha =3,\beta =-12,\gamma =-4$

$\Rightarrow p=\alpha +\beta +\gamma =-13$

Now finding volume of tetrahedron $OABC$

$V=\left|\frac{1}{6}\overrightarrow{OA}·\left(\overrightarrow{OB}\times \overrightarrow{OC}\right)\right|=\left|\frac{1}{6}3\hat{i}·\left(\left(-12\hat{j}\right)\times \left(-4\hat{k}\right)\right)\right|$

$\Rightarrow V=\left|\frac{1}{6}3\hat{i}·\left(48\hat{i}\right)\right|=8\times 3=24$

So, $\left(V,p\right)=\left(24,-13\right)$