If $f\left(x\right)=x^3+p{x}^2+qx+30$ is divisible by $(x-5)$ and when it is divided by $(x+6)$, the remainder is $\text{'}-396\text{'}$, then

$f\left(x\right)=x^3+p{x}^2+qx+30$

$f\left(5\right)=0$

$\Rightarrow 125+25p+5q+30=0$

$25p+5q=-155$

$5p+q=-31\to \left(1\right)$

Given $f(-6)=-396$

$-216+36p-6q+30=-396$

$36p-6q=-210$

$6p-q=-35\to \left(2\right)$

From $\left(1\right)\&\left(2\right)$

$p=-6$

$q=-1$