111) Find the maximum value of the real number $\lambda$. Such that, as long as point $P$ is inside the acute $\triangle ABC$, $\angle PAB = \angle PBC = \angle PCA$, the rays $AP$, $BP$, $CP$ intersect the circumcircles of $\triangle PBC$, $\triangle PCA$, $\triangle PAB$ at points $A_1$, $B_1$, $C_1$ respectively, then $S_{\triangle B C} + S_{\triangle B_1 C A} + S_{\triangle C_1 A B} \geq \lambda S_{\triangle A B C}$.
Let $\triangle A_{1} \cap BC=A_{2}$, and from $\angle PAB = \angle PBC = \angle PA_{1}C$ we know $AB \parallel A_{1}C$. Thus, $\triangle A_{1}A_{2}C \sim \triangle AA_{2}B$.
$$
\frac{S_{\triangle ABC}}{S_{\triangle ABC}} = \frac{A_{1}A_{2}}{AA_{2}} = \frac{CA_{2}}{BA_{2}}.
$$

Similarly, let $B_{2} = BB_{1} \cap CA$, $C_{2} = C_{1} \cap AB$. Then,
$$
\frac{S_{\triangle A_{1}BC} + S_{\triangle B_{1}CA} + S_{\triangle C_{1}AB}}{S_{\triangle ABC}} = \frac{CA_{2}}{BA_{2}} + \frac{AB_{2}}{CB_{2}} + \frac{BC_{1}}{AC_{1}} \\
\geq \sqrt[3]{\frac{AB_{2}}{B_{2}C} \cdot \frac{CA_{2}}{A_{2}B} \cdot \frac{BC_{1}}{C_{1}A}} = 3.
$$

When $\triangle ABC$ is an equilateral triangle and $P$ is its center,
$$
\angle PAB = \angle PBC = \angle PCA = 30^{\circ},
$$
then $A_{2}B = A_{2}C$, $B_{2}A = B_{2}C$, $C_{2}A = C_{2}B$,

at this point the equality holds.
In summary, $\lambda_{\max} = 3$.