7. Let the dihedral angles between the three lateral faces and the base of a tetrahedron all be $60^{\circ}$. The side lengths of the base are $7, 8, 9$. Then the lateral surface area of the pyramid is
7. $24 \sqrt{5}$ vertex $P$ projects onto the base $ABC$ at point $O$, which is the incenter of $\triangle ABC$. Let $r$ be the inradius of $\triangle ABC$, then the slant height $h = 2r$.
So, $S_{\text{trilateral pyramid}} = \frac{1}{2}(a+b+c) h = r(a+b+c) = 2 S_{\triangle ABC}$
$$
=2 \cdot \frac{1}{4} \sqrt{(7+8+9)(7+8-9)(7-8+9)(-7+8+9)} = 24 \sqrt{5}
$$