Solution c0c9ec1d-490d-47c7-921a-d4a6ea1091ac

Points: 2 pts

Lemma 1: $b\in\{-12,0\}$.

Taking the sum and difference of the first and second equations gives
\[
\begin{aligned}
(b+1)(a+c)&=100+87=187,\\
(b-1)(a-c)&=100-87=13.
\end{aligned}
\]
Thus $b+1\mid187$ and $b-1\mid13$. Since $13$ is prime,
\[
b-1\in\{-13,-1,1,13\}
\quad\Longrightarrow\quad
b\in\{-12,0,2,14\}.
\]
The values $b=2$ and $b=14$ would require $3\mid187$ and $15\mid187$,
respectively, which are false. Hence,
\[
b\in\{-12,0\}.
\]

Lemma 2: $ab+bc+ca=276$.

For each candidate, solve
\[
a+c=\frac{187}{b+1},\qquad a-c=\frac{13}{b-1}.
\]
We obtain
\[
\renewcommand{\arraystretch}{1.25}
\begin{array}{c|r|r|r}
b & a & c & ca+b\\ \hline
-12 & -9 & -8 & 60\\
0 & 87 & 100 & 8700
\end{array}
\]
The third equation requires $ca+b=60$, so only
$(a,b,c)=(-9,-12,-8)$ works. Adding the three original equations,
\[
\begin{aligned}
ab+bc+ca&=100+87+60-(a+b+c)\\
&=247-(-29)=276.
\end{aligned}
\]

Final answer:
276\quad\text{(D)}