I read a problem which was as follows:

Equal volumes of $\pu{1M} \ \ce{H3PO4}$ and $\pu{1M} \ \ce{Na3PO4}$ are mixed. The $\mathrm{pH}$ of the resultant mixture is (along with this $K_\mathrm{a1}, K_\mathrm{a2}, K_\mathrm{a3}$ of the acid were mentioned)

The way I approached this problem was as follows: $K_\mathrm{a1} = \frac{[\ce{H2PO4^-}][\ce{H^+}]}{[\ce{H3PO4}]}$, similarly $K_\mathrm{a2} = \frac{[\ce{HPO4^{2-}}][\ce{H^+}]}{[\ce{H2PO4^-}]}$ and so on...

Which gives me $K_\mathrm{a1}\cdot K_\mathrm{a2}\cdot K_\mathrm{a3} = \frac{[\ce{PO4^{3-}}][\ce{H^+}]^3}{[\ce{H3PO4}]} = [\ce{H^+}]^3$ (since $\ce{[H3PO4] = [Na3PO4]}$). Using this, I calculate the $[\ce{H^+}]$ and so the ph which comes out to be $-\log{\sqrt[3]{K_\mathrm{a1}\cdot K_\mathrm{a2}\cdot K_\mathrm{a3}}}$ which is $\frac{\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2} + \mathrm{p}K_\mathrm{a3}}{3}$.

However, it seems from the answer that the system is forming a buffer and $\mathrm{pH}$ is $\mathrm{p}K_\mathrm{a2}$

What am I doing wrong here and how does the $\mathrm{pH}$ come out to be $\mathrm{p}K_\mathrm{a2}$. Is there any such general result or theory for polyprotic acids?