Use Sylow's first and third theorem to conclude that there is only one subgroup of order $137$.
Use Sylow's second theorem to conclude that the subgroup of order $137$ is normal.
Solution: Let $G$ be a group of order $7 \cdot 11 \cdot 137$. By the first Sylow theorem, there exist subgroups of order $7$, $11$ and $137$, say $H_7$, $H_{11} $, and $H_{137} $, respectively. Also, from the Sylow's third theorem, the number of subgroups of order $137$, say $n_{137} $ must satisfy
\begin{align*}
& n_{137} \equiv 1 \ \mathrm{mod}\ 137 \quad \text{ and } \quad n_{137} \mid 77 \\
\implies & n_{137} = 137k + 1 \quad \text{ and } \quad (137k + 1) \mid 77 \\
\implies & k = 0.
\end{align*}
This implies $n_{137} = 1 $. Now using second Sylow's theorem we conclude that there exists $g \in G$ such that $g^{-1} H_{137} g = H_{137} $, which implies $H_{137}$ is normal and hence $G$ is not simple. Therefore, there does not exist any simple group of order $7 \cdot 11 \cdot 137$.