How to calculate pH of solution made of H3PO4 and Na3PO4?
You have the the following six independent equations in order to specify the unknown concentrations of six species as present in an aqueous solution consisting of $\pu{0.1 M}$ $\ce{H3PO4}$ and $\pu{0.05M}$ $\ce{Na3PO4}$:

\begin{align}
K_1 &amp;= \frac{[\ce{H+}][\ce{H2PO4-}]}{[\ce{H3PO4}]} 
     = 7.5 \times 10^{−3} \tag{1}\\
K_2 &amp;= \frac{[\ce{H+}][\ce{HPO4^2-}]}{[\ce{H2PO4-}]} 
     = 6.2 \times 10^{−8} \tag{2}\\
K_3 &amp;= \frac{[\ce{H+}][\ce{PO4^3-}]}{[\ce{HPO4^2-}]} 
     = 3.6 \times 10^{−13} \tag{3}\\
K_\mathrm{w} 
    &amp;= [\ce{H+}][\ce{OH-}] 
     = 1 \times 10^{−14} \tag{4}\\
0.15 
    &amp;=[\ce{PO4^3-}] + [\ce{HPO4^2-}] + [\ce{HPO4-}] + [\ce{H3PO4}]\tag{5}\\
[\ce{H+}] 
    &amp;= 3[\ce{PO4^3-}] + 2[\ce{HPO4^2-}] + 2[\ce{HPO4-}] + [\ce{OH-}] -0.15\tag{6}
\end{align}