7. In tetrahedron $ABCD$, $AD > AB$, $AD \perp AB$, $AD \perp AC$, $\angle BAC = \frac{\pi}{3}$. Let the areas of $\triangle ADB$, $\triangle ADC$, $\triangle ABC$, and $\triangle BCD$ be $S_{1}$, $S_{2}$, $S_{3}$, and $S_{4}$, respectively, and they satisfy $S_{1} + S_{2} = S_{3} + S_{4}$. Then $\frac{S_{3}}{S_{1}} + \frac{S_{3}}{S_{2}} = $ $\qquad$ .
7. $\frac{3}{2}$.

Let $A D=x, A B=y, A C=z$. Then
$$
\begin{array}{l}
B D=\sqrt{x^{2}+y^{2}}, C D=\sqrt{x^{2}+z^{2}}, \\
B C=\sqrt{y^{2}+z^{2}-y z} .
\end{array}
$$

Thus, $S_{4}^{2}=\frac{1}{16}\left(2\left(B C^{2} \cdot B D^{2}+C D^{2} \cdot B D^{2}+B C^{2} \cdot C D^{2}\right)-\right.$
$$
\begin{array}{l}
\left.\quad B C^{4}-B D^{4}-C D^{4}\right) \\
=\frac{1}{4} x^{2} y^{2}+\frac{1}{4} x^{2} z^{2}+\frac{3}{16} y^{2} z^{2}-\frac{1}{4} x^{2} y z \\
=S_{1}^{2}+S_{2}^{2}+S_{3}^{2}-S_{1} S_{2} \\
\text { Also, } S_{1}+S_{2}=S_{3}+S_{4} \text {, hence, } \\
3 S_{1} S_{2}=2 S_{1} S_{3}+2 S_{2} S_{3} \Rightarrow \frac{S_{3}}{S_{1}}+\frac{S_{3}}{S_{2}}=\frac{3}{2}
\end{array}
$$