Suppose $d \in \mathcal{M}$. Then $d \mid 3 \cdot (2n+3m+13)-(6n+6m-1)=3m+40$ and $d \mid 2(3n+5m+1)-(6n+6m-1)=4m+3$. So $d \mid 4(3m+40)-3(4m+3)=151$ which is prime so $\mathcal{M} \subset \{1,151\}$. On the other hand, choosing $m=n=1$ we find that $\text{gcd}(18,9,11)=1 \in \mathcal{M}$ and choosing $m=37, n=89$ we find that $\text{gcd}(302,453, 755)=151 \in \mathcal{M}$ and hence $\mathcal{M}=\{1,151\}=\{d \in \mathcal{N}: d \mid 151\}$.

I think you can just do by Euclidean algorithm gcd of the first two is also just gcd of 2(2n+3m+13)-(3n+5m+1)=n+m+25, gcd of this and the last thing is 6(n+m+25)-(6n+6m-1)=151. So the set M is either 1 or 151, k can be 1 or 151.