First notice that if you have a triplet $(a,b,c)$, then
$$a|b+c$$
$$b|a+c$$
$$c|a+b$$
If the triplet is ordered increasingly, then, you may show that
$$c=a+b$$
With the new information, you get the triplet $(a,b,a+b)$. Therefore
$$a|2b+a$$
$$b|2a+c$$
which means that we need to have
$$a|2b$$
$$b|2a$$
Having $a<b$ and $b|2a$, show that $b=2a$.
The second part is to show that you cannot have more members in the set, for if you add a new integer $d$, to be the greatest in the set, $(a,b,d)$ will not make a triplet.