24. Let $p$, $q$, $r$ be the three distinct real roots of the polynomial $x^{3}-22 x^{2}+80 x-67$. If there exist $A$, $B$, $C \in \mathbf{R}$, such that
$$
\frac{1}{s^{3}-22 s^{2}+80 s-67}=\frac{A}{s-p}+\frac{B}{s-q}+\frac{C}{s-r},
$$

where $s \notin\{p, q, r\}$, then $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=(\quad)$.
(A) 243
(B) 244
(C) 245
(D) 246
(E) 247
24. B.

Since $p, q, r$ are the three distinct real roots of the polynomial $x^{3}-22 x^{2}+80 x-67$, we have
$$
x^{3}-22 x^{2}+80 x-67=(x-p)(x-q)(x-r) \text {. }
$$

By Vieta's formulas, we know
$$
\begin{array}{l}
p+q+r=22, \quad p q+q r+r p=80 . \\
\text { Then } \frac{1}{s^{3}-22 s^{2}+80 s-67} \\
=\frac{1}{(s-p)(s-q)(s-r)}=\frac{A}{s-p}+\frac{B}{s-q}+\frac{C}{s-r} \\
=\frac{A(s-q)(s-r)+B(s-p)(s-r)+C(s-p)(s-q)}{(s-p)(s-q)(s-r)} .
\end{array}
$$

Thus, $A(s-q)(s-r)+B(s-p)(s-r)+$
$$
C(s-p)(s-q)=1 \text {, }
$$

where $s \in \mathbf{R}, s \notin\{p, q, r\}$.
When $s=p$, $A=\frac{1}{(p-q)(p-r)}$.
Similarly,
$$
\begin{array}{l} 
B=\frac{1}{(q-r)(q-p)}, \quad C=\frac{1}{(r-p)(r-q)} . \\
\text { Then } \frac{1}{A}+\frac{1}{B}+\frac{1}{C} \\
=(p-q)(p-r)+(q-r)(q-p)+ \\
(r-p)(r-q) \\
= p^{2}+q^{2}+r^{2}-p q-q r-r p \\
=(p+q+r)^{2}-3(p q+q r+r p) \\
= 244 .
\end{array}
$$