jmc

We can add together the number of lineups with one triplet and with no triplets. The number of lineups with no triplets is $\left(\genfrac{}{}{0ex}{}{11}{6}\right)=462$, since we must choose 6 starters from the 11 remaining players. When one triplet is in the lineup, there are $3\cdot \left(\genfrac{}{}{0ex}{}{11}{5}\right)=1386$ options. So the total number of lineups with at most one triplet is $1386+462=\boxed{1848}$.