In $\triangle A B C$, extend $A B$, $C A$ so that $B A^{\prime}=\lambda_{1} A B, A C^{\prime}=\lambda_{3} C A$, and extend $B C$ so that $C B^{\prime}=\lambda_{2} B C$,
find $S \triangle A^{\prime} B^{\prime} C^{\prime}: S \triangle \triangle B C$.
\begin{array}{l}\text { Similarly, } S \triangle \triangle_{A^{\prime} B B^{\prime}}=\left(1-\lambda_{2}\right) \lambda_{1} S \triangle A B C, \\ S \triangle B^{\prime} B C^{\prime}=\left(1-\lambda_{2}\right)\left(1+\lambda_{3}\right) S \triangle \triangle B C C \\ \text { Also, } S \triangle \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{C}^{\prime}=S \triangle \triangle^{\prime} \mathrm{BC} C^{\prime}+S \triangle \mathrm{A}^{\prime} \mathrm{B}^{\prime} \text { H } \\ +S \triangle \mathrm{B}^{\prime} \mathrm{BC}^{\prime}, \\ \text { Therefore, } \frac{S \triangle \triangle \prime B^{\prime} C^{\prime}}{S \triangle \triangle B C}=1+\left(\lambda_{1}-\lambda_{2}+\lambda_{3}\right) \\ +\left(-\lambda_{1} \lambda_{2}-\lambda_{2} \lambda_{3}+\lambda_{3} \lambda_{1}\right) . \\\end{array}