= Solution
Take base-ten <logarithms> and use dimensionless <thermodynamic activities>. The supplied concentration notation is interpreted in the ideal dilute-solution approximation, with a consistent reference concentration. Put
$$
x=\log_{10}a_{\mathrm{H_4SiO_4}},\qquad
y=\log_{10}(a_{\mathrm{Na^+}}/a_{\mathrm{H^+}}).
$$
With pure solids and unit water activity, the <reaction quotients> are $Q_I=10^{2y+2x}$ and $Q_{II}=10^{2y+6x}$. At <chemical equilibrium>, $Q=K$, so the two lines on the <mineral activity diagram> are
$$
\boxed{I:\ y=3.44-x,\qquad II:\ y=-7.59-3x.}
$$
They have slopes $-1$ and $-3$. Restricting the plotted silica activity to $x<0$ is compatible with their intersection,
$$
x_*=-5.515,\qquad y_*=8.955.
$$
Their full straight extensions are equilibrium relations for the respective phase pairs; not every extension is a stable boundary when the third mineral is available.
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