This one is a little confusing because there is so much symmetry, so let's try to take advantage of that.

Note that the order of the solutions doesn't matter, so to start assume $a \leq b \leq c$. Then consider what the mod statements say:

1. $a \equiv b \pmod{c}$ so $b-a$ is a multiple of $c$.
2. $b\equiv c \pmod{a}$ so $c-b$ is a multiple of $a$.
3. $c\equiv a \pmod{b}$ so $c-a$ is a multiple of $b$.

Note since we assumed $a \leq b \leq c$ we know all the differences ($b-a$, $c-b$, $c-a$) are non-negative. As a hint, think about 1. above. How can $b-a$ be a multiple of $c$ if both $b$ and $a$ are less than $c$? This should get you started.