6 Three distinct positive integers $a$, $b$, and $c$ form a geometric sequence, and their sum is 111. Then $\{a, b, c\}=$ $\qquad$
6 $\{1,10,100\},\{27,36,48\}$ Hint: Let $a<b<c$, and their common ratio be $q$, then $b=a q, c=a q^{2}, q=\frac{b}{a}$ is a rational number, convert it to a reduced fraction, set $q=$ $\frac{n}{m},\left(m<n\right.$, and $m, n$ are coprime), from $c=a \cdot \frac{n^{2}}{m^{2}}$ being a positive integer, so $m^{2} \mid a$, set $a=k m^{2}$, then
$111=1 \cdot 111=3 \cdot 37=a+b+c=k\left(m^{2}+m n+n^{2}\right)$.
When $k=1$, $m^{2}+m n+n^{2}=111, n \leqslant 10$, we get one solution $m=1, n=10$; $\{a, b, c\}=\{1,10,100\}$.
When $k=3$, $m^{2}+m n+n^{2}=37, n \leqslant 6$, we get another solution $m=3, n=4$; $\{a, b, c\}=\{27,36,48\}$.