jmc

We can break this problem into three parts. First, we will count the number of ways to choose teachers for the subcommittee. We need to choose 2 teachers from the 5 in the school organization, which gives $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ ways to choose the teachers. Next, we choose parents. We need to choose 3 parents out of 7, which gives $\left(\genfrac{}{}{0ex}{}{7}{3}\right)=35$ ways to choose the parents. Finally, we choose the students. There are 3 students chosen from 6, which gives $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$ ways to choose the students. Each of these selections is independent, so the total number of possible subcommittees will be $10\cdot 35\cdot 20=\boxed{7000}$.