Romanian Mathematical Olympiad

If one of the numbers is $0$, then all are $0$.
If $abc \neq 0$, then the relation can be written $\frac{1}{bc} + \frac{1}{ac} + \frac{1}{ab} = 1$. Since the relation is symmetric in $a, b, c$, we may assume that $a \le b \le c$, whence $ab \le ac \le bc$.
If $ab > 3$, then $\frac{1}{bc} + \frac{1}{ac} + \frac{1}{ab} < 1$.
If $ab = 2$, then $a = 1$ and $b = 2$, whence $\frac{1}{2c} + \frac{1}{c} = \frac{1}{2}$, so $c = 3$.

If $ab = 3$, then $a = 1$ and $b = 3$, whence $\frac{1}{3c} + \frac{1}{c} = \frac{2}{3}$, so $c = 2$, which contradicts $b < c$.
Finally, the solutions are $(0, 0, 0)$ and all the permutations of $(1, 2, 3)$.