Given $P$ is a point inside $\triangle A B C$, and the rays $A P, B P, C P$ intersect the sides $B C, C A, A B$ at points $D, E, F$ respectively. If $S_{\triangle A B C}=\lambda S_{\triangle O E F}$, find the value of $\frac{P A}{P D}+\frac{P B}{P E}+\frac{P C}{P F}$ (expressed in terms of $\lambda$).
As shown in Figure 2, let $E F$ intersect $A P$ at point $H$. By the area ratio, we have
$$
\begin{array}{l}
\frac{A H}{P H}=\frac{S_{\triangle A E F}}{S_{\triangle P E F}} \\
=\frac{S_{\triangle A E F}}{S_{\triangle A P F}} \cdot \frac{S_{\triangle A P F}}{S_{\triangle P E F}} \\
=\frac{E B}{P B} \cdot \frac{A C}{E C} \\
=\frac{S_{\triangle E B C}}{S_{\triangle P B C}} \cdot \frac{S_{\triangle A B C}}{S_{\triangle E B C}}=\frac{S_{\triangle A B C}}{S_{\triangle P B C}}=\frac{A D}{P D} .
\end{array}
$$

Thus, $A D \cdot P H=A H \cdot P D$
$$
\begin{array}{l}
=(A P-P H)(A D-A P) \\
=A P(A D-A P+P H)-A D \cdot P H \\
=A P \cdot D H-A D \cdot P H \\
\Rightarrow A P \cdot D H=2 A D \cdot P H \\
\Rightarrow \frac{P H}{D H}=\frac{P A}{2 A D} . \\
\quad \\
\quad= \frac{A D}{P D} \cdot \frac{P B \cdot P C}{P E \cdot P F} \cdot \frac{P H}{D H} \\
= \frac{P A \cdot P B \cdot P C}{2 P D \cdot P E \cdot P F} .
\end{array}
$$

Let $S_{\triangle P B C}=x, S_{\triangle P C A}=y, S_{\triangle P A B}=z$. Then
$$
\frac{P A}{P D}=\frac{z}{S_{\triangle P B D}}=\frac{y}{S_{\triangle P C D}}=\frac{z+y}{x} \text {. }
$$

Similarly, $\frac{P B}{P E}=\frac{x+z}{y}, \frac{P C}{P F}=\frac{y+x}{2}$.
Thus, $2 \lambda=\frac{P A}{P D} \cdot \frac{P B}{P E} \cdot \frac{P C}{P F}$
$$
\begin{aligned}
= & \frac{(x+y)(y+z)(z+x)}{x y z} \\
= & \frac{1}{x y z}[y z(z+y)+z x(z+x)+ \\
& x y(x+y)+2 x y z] \\
= & \frac{z+y}{x}+\frac{x+z}{y}+\frac{y+x}{z}+2 \\
= & \frac{P A}{P D}+\frac{P B}{P E}+\frac{P C}{P F}+2 .
\end{aligned}
$$

Therefore, $\frac{P A}{P D}+\frac{P E}{P E^{\prime}}+\frac{P C}{P F}=2 \lambda-2$.