6. Given three distinct positive integers $a, b, c$ form a geometric sequence, and their sum is 111. Then $\{a, b, c\}=$ $\qquad$
6. $\{1,10,100\},\{27,36,48\}$.

Let $a<b<c$ and the common ratio be $q$. Then $b=a q, c=a q^{2} \Rightarrow q=\frac{b}{a}$ is a rational number.
Express $q$ as a reduced fraction, denoted as $q=\frac{n}{m}$, where $m<n$, and $(m, n)=1$.
Since $c=a \cdot \frac{n^{2}}{m^{2}}$ is a positive integer $\Rightarrow m^{2} \mid a$.
Let $a=\mathrm{km}^{2}$. Then
$$
\begin{array}{l}
111=1 \times 111=3 \times 37 \\
=a+b+c=k\left(m^{2}+m n+n^{2}\right) .
\end{array}
$$

When $k=1$,
$$
\begin{array}{l}
m^{2}+m n+n^{2}=111 \\
\Rightarrow n \leqslant 10, m=1, n=10 \\
\Rightarrow\{a, b, c\}=\{1,10,100\} ;
\end{array}
$$

When $k=3$,
$$
\begin{array}{l}
m^{2}+m n+n^{2}=37 \\
\Rightarrow n \leqslant 6, m=3, n=4 \\
\Rightarrow\{a, b, c\}=\{27,36,48\} .
\end{array}
$$