Example 4 As shown in Figure $4, S_{\triangle P B_{2} A_{1}}=3, S_{\triangle P A_{1} A_{2}}=2$, $S_{\triangle P A_{2} B_{1}}=1$. Find $S_{\triangle A_{3} B_{2} B_{1}}$.
(1990, Shanghai Junior High School Mathematics Competition)
Solution: With the symbols having the same meanings as before, we have
$$
\begin{array}{l}
\lambda_{3}=S_{\triangle P A_{1} A_{2}}=2, \\
S_{\triangle P B_{2} A_{1}}=\frac{A_{1} B_{2}}{A_{1} A_{3}} \\
\cdot S_{\triangle P A_{3} A_{1}} \\
=\frac{\lambda_{3}}{\lambda_{3}+\lambda_{1}} \cdot \lambda_{2} \\
=3, \\
S_{\triangle P A_{2} B_{1}}=\frac{A_{2} B_{1}}{A_{2} A_{3}} \cdot S_{\triangle P A_{2} A_{3}} \\
=\frac{\lambda_{3}}{\lambda_{2}+\lambda_{3}} \cdot \lambda_{1}=1 .
\end{array}
$$

Solving, we get $\lambda_{1}=S_{\triangle P A_{2} A_{3}}=10, \lambda_{2}=S_{\triangle P A_{3} A_{1}}=18$. Then from (1) and (2), we have
$$
\begin{array}{c}
\mu_{3}=\frac{\lambda_{1} \lambda_{2} \lambda_{3}}{\left(\lambda_{2}+\lambda_{3}\right)\left(\lambda_{3}+\lambda_{1}\right)}=\frac{3}{2}, \\
v_{3}=\frac{\mu_{3}}{\lambda_{3}}\left(\lambda_{1}+\lambda_{2}+\lambda_{3}\right)=\frac{45}{2},
\end{array}
$$

which means $S_{\triangle A_{3} B_{1} B_{2}}=\frac{45}{2}$.