6. In the triangular pyramid $P-ABC$, the angles formed by the three lateral faces and the base are equal, the areas of the three lateral faces are $3$, $4$, and $5$, and the area of the base is $6$. Then, the surface area of the circumscribed sphere of the triangular pyramid $P-ABC$ is $\qquad$
6. $\frac{79 \pi}{3}$.

Let the angle between the side faces and the base be $\theta$. By the projection theorem of area,
$$
\cos \theta=\frac{6}{3+4+5}=\frac{1}{2} \Rightarrow \theta=60^{\circ} .
$$

It is easy to see that the projection of $P$ on the base is exactly the incenter of $\triangle A B C$, denoted as $I$.

Since the areas of the three side faces are $3$, $4$, and $5$, the ratio of the sides of $\triangle A B C$ is $3: 4: 5$.

Noting that the area of $\triangle A B C$ is 6, we know that the inradius of $\triangle A B C$ is 1.
Thus, the height of the tetrahedron $P-A B C$ is $\sqrt{3}$.
Let the circumcenter of the tetrahedron $P-A B C$ be $O$ and the radius be $R$. Since $\triangle A B C$ is a right triangle with side lengths $3$, $4$, and $5$, the distance between the incenter $I$ and the circumcenter $O'$ of $\triangle A B C$ is $\frac{\sqrt{5}}{2}$. Since the sphere center is outside the tetrahedron $P-A B C$, constructing a right triangle easily gives
$$
\begin{array}{l}
\left(\sqrt{3}+\sqrt{R^{2}-\frac{25}{4}}\right)^{2}+\left(\frac{\sqrt{5}}{2}\right)^{2}=R^{2} \\
\Rightarrow R^{2}=\frac{79}{12} .
\end{array}
$$

Therefore, the surface area of the circumscribed sphere $S=4 \pi R^{2}=\frac{79 \pi}{3}$.