7. In the tetrahedron $P-ABC$, it is known that $AB=\sqrt{3}, AC=1, PB=PC=\sqrt{2}$, then the range of $S_{\triangle ABC}^{2}+S_{\triangle PBC}^{2}$ is $\qquad$
The answer is $\left(\frac{1}{4}, \frac{7}{4}\right]$.
Analysis: When $S_{\triangle A B C}^{2}+S_{\triangle P B C}^{2}>S_{\triangle P B C}^{2} \geq \frac{1}{4}$ and $B C=2$, $A B \perp A C$, $P B \perp P C$, the maximum value is achieved. $S_{\triangle A B C}^{2}+S_{\triangle P B C}^{2} \leq \frac{3}{4}+1=\frac{7}{4}$, so the range of values is $\left(\frac{1}{4}, \frac{7}{4}\right]$.