Select the first junior and call this person $A$. There are 5 other juniors and $\binom{6}{2} = 15$ pairs of seniors who could team up with $A$. Select the next junior not yet on a team, say $B$. There are 3 other juniors and $\binom{4}{2} = 6$ pairs of seniors who could team up with $B$. The third team consists of the people not yet chosen. Thus there are $5 \cdot 15 \cdot 3 \cdot 6 = 1350$ ways to form the teams.

There are $\binom{6}{2,2,2} = \frac{6!}{2!2!2!} = 90$ ways to assign the seniors to teams 1, 2, and 3, and, similarly, 90 ways to assign juniors to those teams. But there are $3! = 6$ ways for the three teams to be ordered, so the number of ways to form the teams is $\frac{90 \cdot 90}{6} = 1350$.