2019 AMC 10A Problem 24

Multiplying the identity by $(s-p)(s-q)(s-r)$ gives $$\begin{aligned} 1={}&A(s-q)(s-r)\\ &+B(s-p)(s-r)\\ &+C(s-p)(s-q). \end{aligned}$$ Setting $s=p,q,r$, in turn, yields $$\begin{aligned} \frac1A&=(p-q)(p-r),\\ \frac1B&=(q-p)(q-r),\\ \frac1C&=(r-p)(r-q). \end{aligned}$$ Adding and expanding gives $$\begin{aligned} \frac1A+\frac1B+\frac1C &=p^2+q^2+r^2\\ &\quad-pq-pr-qr. \end{aligned}$$ By Vieta’s formulas, $p+q+r=22$ and $pq+pr+qr=80$, so $$\begin{aligned} p^2+q^2+r^2&=22^2-2(80)\\ &=324. \end{aligned}$$ Therefore the requested value is $324-80=244.$ Thus, B is the correct answer.