3 lines are to divide up 9 wrenches evenly.

$\displaystyle \displaystyle\frac{9!}{(3!)^3}=1680$

I have this part correct but I can't seem to get this next part.

If there are 2 used and 7 new wrenches, what is the probability that a particular line (line A) gets both used?

Here the answer is 1/12