22. If $a, b, c$ satisfy the conditions, then for any $k \geqslant 1, k a, k b, k c$ also satisfy them, so we can first assume $(a, b, c)=1$, and $1 \leqslant a \leqslant b \leqslant c$. From this and $c \mid a-b$ we deduce $a=b$, and then by $a \mid c$ and $(a, b, c)=1$ we deduce $a=b=1$. Therefore, all solutions are $\{1,1, c\}, c$ being any positive integer.