Solution: $60+12 \sqrt{5}$.
By Heron's formula, it is easy to know that $S_{\triangle A B C}=12 \sqrt{5}$. As shown in the figure, construct $H M \perp B C, H N \perp C A, H K \perp A B$, with $M, N, K$ being the feet of the perpendiculars. By the three perpendiculars theorem, we know that $D M \perp B C, D N \perp C A, D K \perp A B$. Let $r$ be the inradius of $\triangle A B C$, then $R t \triangle D H M, R t \triangle D H N, R t \triangle D H K$ are congruent. Denote
$$
D M=D N=D K=t, D H=h .
$$

From $r(7+8+9)=2 S_{\triangle A B C}=24 \sqrt{5}$, we get $r=\sqrt{5}$.
And from the volume $40=V=\frac{1}{3} h S_{\triangle A B C}=4 \sqrt{5} h$, we get $h=2 \sqrt{5}$.
Thus, $t=\sqrt{r^{2}+h^{2}}=5$.
Therefore, $S_{\triangle A B D}+S_{\triangle B C D}+S_{\triangle C A D}=\frac{t}{2}(A B+B C+C A)=60$.
Hence, $S_{\text {surface area }}=60+12 \sqrt{5}$.
7. Given a tetrahedron \(ABCD\) with a volume of \(40\), \(AB = 7\), \(BC = 8\), \(AC = 9\). If the projection of vertex \(D\) onto the plane \(ABC\) is exactly the incenter \(H\) of \(\triangle ABC\), then the surface area of the tetrahedron is

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