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 $$ \frac{1}{A}+\frac{1}{B}+\frac{1}{C}=p^{2}+q^{2}+r^{2}-p q-q r-p r $$ By Vieta\textquotesingle s Formulas, we know that $p^{2}+q^{2}+r^{2}=(p+q+r)^{2}-2(p q+q r+p r)=324$ and $p q+q r+p r=80$. Thus the answer is $324-80=$ 244 . 
Let $p, q$, and $r$ be the distinct roots of the polynomial $x^{3}-22 x^{2}+80 x-67$. It is given that there exist real numbers $A, B$, and $C$ such that $$ \frac{1}{s^{3}-22 s^{2}+80 s-67}=\frac{A}{s-p}+\frac{B}{s-q}+\frac{C}{s-r} $$ for all $s 
otin\textbackslash{p, q, r\textbackslash}$. What is $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}$? 
Options:
[1] $$243$$ 
[2] $$244$$ 
[3] $$245$$ 
[4] $$246$$ 
[5] $$247$$ 