imc

Multiplying both sides by $(s-p)(s-q)(s-r)$ yields $$1=A(s-q)(s-r)+B(s-p)(s-r)+C(s-p)(s-q)$$ As this is a polynomial identity, and it is true for infinitely many $s$, it must be true for all $s$ (since a polynomial with infinitely many roots must in fact be the constant polynomial $0$). This means we can plug in $s=p$ to find that $\frac{1}{A}=(p-q)(p-r)$. Similarly, we can find $\frac{1}{B}=(q-p)(q-r)$ and $\frac{1}{C}=(r-p)(r-q)$. Summing them up, we get that $$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=p^2+q^2+r^2-pq-qr-pr$$ We can express $p^2+q^2+r^2=(p+q+r{)}^2-2(pq+qr+pr)$, and by Vieta's Formulas, we know that this expression is equal to $484$. Vieta's also gives $pq+qr+pr=80$ (which we also used to find $p^2+q^2+r^2$), so the answer is $484-240=\boxed{244}$. Note: this process of substituting in the 'forbidden' values in the original identity is a standard technique for partial fraction decomposition, as taught in calculus classes.