Define a polynomial $f(x) = (x-a)(x-b)(x-c)$

Let $d = abc-(a+b+c) = (ab+bc+ac)-1$ . Note $d > 0$ . Then $(a+i)(b+i)(c+i) = d+di$ so the complex argument of the product is $45^o$ , and the arguments of the terms are between $0$ and $45^o$ , and the arguments of the terms are in decreasing order and sum to $45^o$ , so arg $(a+i)$ is at least $15^o$ , so $a \le \frac1{\tan(15^o)} \approx 3.732$ , so $a =1,2,3$ are the only possibilities.

$a = 1$ simplifies to $b+c = 0$ , which is impossible.

$a = 2$ simplifies to $(b-3)(c-3)=10$ , which leads to $(2,4,13)$ and $(2,5,8)$ .

$a = 3$ simplifies to $(b-2)(c-2) = 5$ , which leads to $(3,3,7)$ .

So there are a total of $\fbox{3}$ solutions.

There is probably a way of getting the inequality without using complex analysis, but I think this way is pretty clean.