Inspired by Abner Chinga

Observe $(a,b,c) = (1,1,-1)$ . This is a solution, but it refutes three of the four choices: $abc = -1 \neq 0$ , $ab+bc+ca = -1 \neq 0$ , and $a+b = 2 \neq 0$ . Thus in order for the problem to have exactly one answer, the remaining choice $(a+b)(b+c)(c+a) = 0$ must be the correct answer. We need to prove that this is indeed correct.

$\displaystyle\begin{aligned} \frac{1}{a}+\frac{1}{b}+\frac{1}{c} &= \frac{1}{a+b+c} \\ \frac{1}{a}+\frac{1}{b} &= \frac{1}{a+b+c}-\frac{1}{c} \\ \frac{a+b}{ab} &= \frac{-(a+b)}{(a+b+c)c} \end{aligned}$ $\displaystyle\begin{aligned} (a+b) \cdot \left( \frac{1}{ab} + \frac{1}{(a+b+c)c} \right) &= 0 \\ (a+b) \cdot \left( \frac{(a+b+c)c+ab}{abc(a+b+c)} \right) &= 0 \\ (a+b) \cdot \left( \frac{(a+c)(b+c)}{abc(a+b+c)} \right) &= 0 \\ (a+b)(a+c)(b+c) \cdot \frac{1}{abc(a+b+c)} &= 0 \end{aligned}$

Since $a,b,c,a+b+c \neq 0$ (otherwise the equation in the problem is not defined), $\frac{1}{abc(a+b+c)} \neq 0$ , thus $(a+b)(a+c)(b+c) = 0$ .