Example 9 In $\triangle A B C$, $O$ is the intersection of $A A^{\prime}$ and $B B^{\prime}$, $A^{\prime}$ is on $B C$, $B^{\prime}$ is on $A C$, and $S_{\triangle A O R^{\prime}}=3, S_{\triangle O A B}=2, S_{\triangle O A^{\prime} B}=1$. Find $S_{\triangle C A^{\prime} B^{\prime}}$.
Notice the conclusions 2 and 3, and the symbols have the same meanings as above, we have
$$
\begin{array}{c}
\lambda_{3}=2 \\
S_{\triangle A O B^{\prime}}=\frac{A B^{\prime}}{A C} \cdot S_{\triangle O A C}=\frac{\lambda_{3}}{\lambda_{3}+\lambda_{1}} \cdot \lambda_{2}=3 \\
S_{\triangle O A^{\prime} B}=\frac{A^{\prime} B}{B C} \cdot S_{\triangle O B C}=\frac{\lambda_{3}}{\lambda_{2}+\lambda_{3}} \cdot \lambda_{1}=1
\end{array}
$$

From the above, we get
$$
\lambda_{1}=S_{\triangle O B C}=10, \lambda_{2}=S_{\triangle O A C}=18
$$

Thus,
$$
\begin{aligned}
\mu_{3} & =\frac{\lambda_{1} \lambda_{2} \lambda_{3}}{\left(\lambda_{2}+\lambda_{3}\right)\left(\lambda_{3}+\lambda_{1}\right)}=\frac{3}{2} \\
v_{3} & =\frac{\mu_{3}}{\lambda_{3}}\left(\lambda_{1}+\lambda_{2}+\lambda_{3}\right)=\frac{45}{2}
\end{aligned}
$$

Therefore, $S_{\triangle A^{\prime} B^{\prime} C}=\frac{45}{2}$.