Find the largest value of the real number $ \lambda$, such that as long as point $ P$ lies in the acute triangle $ ABC$ satisfying $ \angle{PAB}\equal{}\angle{PBC}\equal{}\angle{PCA}$, and rays $ AP$, $ BP$, $ CP$ intersect the circumcircle of triangles $ PBC$, $ PCA$, $ PAB$ at points $ A_1$, $ B_1$, $ C_1$ respectively, then $ S_{A_1BC}\plus{} S_{B_1CA}\plus{} S_{C_1AB} \geq \lambda S_{ABC}$.
1. **Define the angles and setup the problem:**
   Let $\angle PAB = \angle PBC = \angle PCA = \omega$. Since $BPCA_1$ is cyclic, we have $\angle AA_1C = \angle PBC = \omega$. This implies that $CA_1 \parallel AB$.

2. **Relate areas using parallel lines:**
   Since $CA_1 \parallel AB$, the triangles $A_1BC$ and $ABC$ are similar. Therefore, the ratio of their areas is given by the square of the ratio of corresponding sides:
   \[
   \frac{S_{A_1BC}}{S_{ABC}} = \left(\frac{CA_1}{AB}\right)^2
   \]

3. **Express the ratio of areas in terms of trigonometric functions:**
   Using the fact that $CA_1 \parallel AB$, we can write:
   \[
   \frac{S_{A_1BC}}{S_{ABC}} = \frac{AM}{MA_1} = \frac{MC}{MB} = \frac{S_{PAC}}{S_{PAB}}
   \]
   where $M = BC \cap AA_1$. Hence,
   \[
   S_{A_1BC} = \frac{S_{PAC}}{S_{PAB}} \cdot S_{ABC} = \frac{PA \cdot PC \cdot \sin A}{PA \cdot PB \cdot \sin B} \cdot S_{ABC} = \frac{\sin \omega \cdot \sin A}{\sin (C - \omega) \cdot \sin B} \cdot S_{ABC}
   \]

4. **Simplify the expression using the given angle condition:**
   Given $\omega = \angle PAB = \angle PBC = \angle PCA$, we use the identity $\cot \omega = \cot A + \cot B + \cot C$. This leads to:
   \[
   S_{A_1BC} = \frac{\sin^2 A}{\sin^2 C} \cdot S_{ABC} = \frac{a^2}{c^2} \cdot S_{ABC}
   \]

5. **Formulate the inequality:**
   Similarly, we can find:
   \[
   S_{B_1CA} = \frac{c^2}{b^2} \cdot S_{ABC} \quad \text{and} \quad S_{C_1AB} = \frac{b^2}{a^2} \cdot S_{ABC}
   \]
   Therefore, the problem reduces to finding the largest $\lambda$ such that:
   \[
   \frac{a^2}{c^2} + \frac{c^2}{b^2} + \frac{b^2}{a^2} \geq \lambda
   \]

6. **Analyze the inequality:**
   By the AM-GM inequality, we have:
   \[
   \frac{a^2}{c^2} + \frac{c^2}{b^2} + \frac{b^2}{a^2} \geq 3 \sqrt[3]{\frac{a^2}{c^2} \cdot \frac{c^2}{b^2} \cdot \frac{b^2}{a^2}} = 3
   \]
   Hence, $\lambda \geq 3$.

7. **Verify the equality case:**
   The equality holds when $a = b = c$, i.e., when the triangle is equilateral. In this case:
   \[
   \frac{a^2}{c^2} + \frac{c^2}{b^2} + \frac{b^2}{a^2} = 3
   \]

Conclusion:
\[
\boxed{\lambda = 3}
\]