Feynman diagram for the decay $\tau \rightarrow \nu_{\tau}\pi^+\pi^-\pi^-$

$$\tau \rightarrow \nu_{\tau}\pi^+\pi^-\pi^-$$

This is a possible decay mode of the $$\tau$$ lepton (see Wikipedia). Given that a neutrino is present this must be the weak interaction and therefore we can imagine the Feynman diagram will look something like this:

What would we draw in place of "pions"? Is it possible for a W boson to produce 6 quarks? I don't see why not, but this is not something I have ever seen drawn before. If not, through what interaction would all 6 quarks be produced at this vertex?

The $$W^-$$ gives a $$d \overline u$$ quark-antiquark pair. That's the only relevant diagram: extra $$W$$, $$Z$$ and $$\gamma$$ lines push the rate right down.
For this particular $$\tau$$ decay the quark and antiquark have their spins aligned, $$S=1$$, they have orbital angular momentum $$L=1$$, and the two combine to give $$J=1$$. This gives the $$a_1$$ meson (at least, that's the dominant process). The $$a_1$$ has a mass of 1260 MeV and a width of about 450 MeV (so it's very broad). It decays to 3 pions through $$\rho \pi$$ intermediate states. This is a low energy strong process and can't really be described in terms of gluons (it would need an infinite number of diagrams in a non-converging series).