The question is ....

We have that

(I) $a + b \equiv 0\mod{c} \Longrightarrow a + b + c \equiv 0\mod{c}$ ,

(II) $a + c \equiv 0\mod{b} \Longrightarrow a + b + c \equiv 0\mod{b}$ , and

(III) $b + c \equiv 0\mod{a} \Longrightarrow a + b + c \equiv 0\mod{a}$ .

These imply that $a + b + c \equiv 0 \mod{abc} \Longrightarrow a + b + c \ge abc$ .

Now look at the options for $a$ . If $a = 1$ then we have that

$1 + b + c \ge bc \Longrightarrow b \le \frac{c + 1}{c - 1}$ for $c \ne 1$ .

For $a = b = c = 1$ we have all of conditions (i), (ii) and (iii) being met, so $(1,1,1)$ is one solution triple. If $c = 2$ then we would require, (with $a = 1$ ), that $b \le \frac{3 + 1}{3 - 1} = 3$ , but since we also require that $b \le c$ this limit us to either $b = 1$ or $b = 2$ . Now the triple $(1,1,2)$ satisfies all of (i), (ii) and (iii), and is thus a solution triple, but $(1,2,2)$ does not satisfy condition (iii).

If $c = 3$ then $b \le \frac{3 + 1}{3 - 1} = 2$ . This gives us the possible solution triples $(1,1,3)$ and $(1,2,3)$ , only the second of which satisfies all of the original conditions.

So for $a = 1$ we have the $3$ solution triples $(1,1,1), (1,1,2)$ and $(1,2,3)$ .

If $a = 2$ then $2 + b + c \le 2bc \Longrightarrow c \le \frac{b + 2}{2b - 1}$ . Now since we must have $a \le b$ we start by looking at the case $b = 2$ . This implies that $c \le \frac{4}{3}$ , i.e., that $c = 1$ , since $c$ must be an integer. But this violates the condition that $b \le c$ , and so $b \ne 2$ . For $b = 3$ we would require that $c \le 1$ , again violating the condition that $b \le c$ . For $b \gt 3$ we would require that $c \lt 1$ , which has no positive integer solution.

For $a = 3$ we would require that $c \le \frac{b + 3}{3b - 1}$ , which, since we would require that $b \ge 3$ , would require that $c \lt 1$ . This will be the case for any $a \ge 3$ , so the only solution triples are those we found with $a = 1$ .

We thus conclude that there are $\boxed{3}$ solution triples.