Answer (D): Notice that the difference between $100$ and $87$ is $13$, a prime number. This fact will help to simplify the problem. Subtract the second equation from the first to get
$$\begin{aligned} 13 &= (ab + c) - (bc + a) \\ &= ab - bc - a + c \\ &= b(a - c) - (a - c) \\ &= (b - 1)(a - c). \end{aligned}$$
Thus $b - 1 = \pm 1$ or $b - 1 = \pm 13$.

* If $b - 1 = -1$, then $b = 0$, implying $c = 100$, $a = 87$, and $ca = 60$, which is impossible.
* If $b - 1 = 1$, then $b = 2$ and $a - c = 13$, implying $ca = 58 = 2 \cdot 29$, which cannot be true if $a - c = 13$.
* If $b - 1 = 13$, then $b = 14$ and $a - c = 1$, implying $ca = 46 = 2 \cdot 23$, which cannot be true if $a - c = 1$.
* If $b - 1 = -13$, then $b = -12$ and $a - c = -1$, implying $ca = 72$, which is satisfied when $a = -9$ and $c = -8$. In fact, $a = -9$, $b = -12$, and $c = -8$ satisfies all three equations.

The requested value is $ab + bc + ca = (-9)(-12) + (-12)(-8) + (-8)(-9) = 108 + 96 + 72 = 276$.