Solve the equation $1 / a+1 / b+1 / c=1$ in integers.

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If $a=1$, then $c=-b$.

Let none of the numbers be equal to 1. Then each of the terms is no more than $1 / 2$. From this, it immediately follows that none of the numbers can be negative. Therefore, we can assume that if $13$, then $1 / a+1 / b+1 / c<1 / 3+1 / 3+1 / 3=1$.

Therefore, only two cases are possible:

1) $a=3$ (and then, obviously, $b=c=3$);
2) $a=2$. In this case, $1 / b+1 / c=1 / 2$. Here, $b \leq 4$.

Therefore, either $b=4$ (and then $c=4$), or $b=3$ (and then $c=6$).

## Answer

$(3,3,3),\{2,3,6\},\{2,4,4\},\{1, t,-t\}$, where $t-$ is any integer.