7. Let a fixed point $P$ outside the plane $\alpha$ be at a distance $h$ from $\alpha$, and let three moving points $A, B, C$ on $\alpha$ be at distances $a, b, c$ from $P$, respectively, with $\angle PBA = 90^{\circ}$. Then the maximum area of $\triangle ABC$ is $\qquad$ (expressed in terms of $a, b, c, h$).
7. $\frac{1}{2} \sqrt{a^{2}-b^{2}}\left(\sqrt{b^{2}-h^{2}}+\sqrt{c^{2}-h^{2}}\right)$.

Notice that $|A B|=\sqrt{a^{2}-b^{2}}$.
Let the projection of $P$ on plane $\alpha$ be $Q$. Then $A B \perp B Q$.
Thus, $|B Q|=\sqrt{b^{2}-h^{2}},|C Q|=\sqrt{c^{2}-h^{2}}$.
Therefore, the distance from point $C$ to $A B$ is less than or equal to $|B Q|+|C Q|$.