6. As shown in the figure, in the tetrahedron $P-ABC$, $\triangle PAC$ and $\triangle ABC$ are both equilateral triangles with side lengths of 6. If the dihedral angle $P-AC-B$ is $120^{\circ}$, then the surface area of the circumscribed sphere of the tetrahedron $P-ABC$ is $\qquad$.
Answer $84 \pi$.
Analysis As shown in the figure, take the midpoint $D$ of $AC$, and connect $DP, DB$.
Since $\triangle PAC$ and $\triangle ABC$ are both equilateral triangles with side length 6, we have $PD \perp AC, BD \perp AC$, and $\angle PDB$ is the dihedral angle of $P-AC-B$, with $\angle PDB=120^{\circ}$.

Let $O$ be the center of the circumscribed sphere of the tetrahedron $P-ABC$, and $O_{1}, O_{2}$ be the centroids of $\triangle ABC$ and $\triangle PAC$, respectively. Then $OO_{1} \perp$ plane $ABC$, $OO_{2} \perp$ plane $PAC$, and
$$
O_{2}D=O_{1}D=\frac{1}{3}\left(\frac{\sqrt{3}}{2} \times 6\right)=\sqrt{3}, OO_{2}=OO_{1}
$$

It is easy to see that $O, O_{2}, D, O_{1}$ are coplanar. Connect $OD$, then
$$
\angle ODO_{1}=60^{\circ}, OO_{1}=\sqrt{3}DO_{1}=3
$$

Therefore, the radius of the circumscribed sphere of the tetrahedron $P-ABC$ is
$$
R=OB=\sqrt{OO_{1}^{2}+O_{1}B^{2}}=\sqrt{3^{2}+(2\sqrt{3})^{2}}=\sqrt{21}
$$

Thus, the surface area of the circumscribed sphere of the tetrahedron $P-ABC$ is $4 \pi R^{2}=84 \pi$.