3. In the tetrahedron $P-ABC$, $PC \perp$ plane $ABC$, $AB=8$, $BC=6$, $PC=9$, $\angle ABC=120^{\circ}$. Then the cosine value of the dihedral angle $B-AP-C$ is $\qquad$
3. $1 \frac{\sqrt{111}}{148}$.

Draw $C K \perp A B$ extended at $K$, and connect $P K$. We have
$$
\begin{array}{l}
P K \perp A B, C K=6 \sin 60^{\circ}=3 \sqrt{3} . \\
P K=\sqrt{9^{2}+(3 \sqrt{3})^{2}}=6 \sqrt{3}, \\
B K=3 .
\end{array}
$$

Let $\angle P K C=\alpha$. Then
$$
\begin{array}{c}
\cos \alpha=\frac{3 \sqrt{3}}{6 \sqrt{3}}=\frac{1}{2}, \\
\alpha=60^{\circ} . \\
\text { Also, } S_{\triangle A B P}=\frac{1}{2} A B \\
\cdot P K=\frac{1}{2} \times 8 \times 6 \sqrt{3}=24 \sqrt{3} .
\end{array}
$$

Draw $B H \perp A C$ at $H$. Then plane $A P C \perp$ plane $A B C$, $B H \perp$ plane $A P C$. Draw $B Q \perp A P$ at $Q$, and connect $Q H$. Then $Q H \perp$ $A P$. Connect $B H . \angle B Q H$ is the plane angle of the dihedral angle $B-A P-C$, let $\angle B Q H=\beta$.
$$
\begin{array}{l}
A C \cdot A H=A B \cdot A C, \\
A C=\sqrt{11^{2}+(3 \sqrt{3})^{2}}=\sqrt{148}, \\
\therefore \quad A H=\frac{8 \times 11}{\sqrt{148}}=\frac{44}{\sqrt{37}} . \\
S_{\triangle P A H}=\frac{1}{2} \cdot A H \cdot P C \\
=\frac{1}{2} \times 9 \times \frac{44}{\sqrt{37}}=\frac{198}{\sqrt{37}} . \\
\therefore \cos \beta=\frac{S_{\triangle P A H}}{S_{\triangle P A B}}=\frac{\frac{198}{\sqrt{37}}}{24 \sqrt{3}}=\frac{11 \sqrt{111}}{148} .
\end{array}
$$