My solution: Let $ \frac{A'C}{BA'}=\alpha, \frac{B'A}{CB'}=\beta, \frac{C'B}{AC'}=\gamma $ . From Ceva theorem we get $ \alpha \beta \gamma =1 $ . From Van Aubel's theorem we get $ \frac{AO}{OA'}=\beta+\frac{1}{\gamma}, \frac{BO}{OB'}=\gamma+\frac{1}{\alpha}, \frac{CO}{OC'}=\alpha+\frac{1}{\beta} $ , so $ \frac {AO}{OA'}\cdot \frac {BO}{OB'}\cdot \frac {CO}{OC'}=(\beta+\frac{1}{\gamma})(\gamma+\frac{1}{\alpha})(\alpha+\frac{1}{\beta}) =2+\frac {AO}{OA'}+\frac {BO}{OB'}+\frac {CO}{OC'}=94 $ . Q.E.D

Let $[]$ be enclosed area. $[BOC]=a; [COA]=b; [AOB]=c \Rightarrow \frac{AO}{OA'}=\frac{b+c}{a} \Rightarrow \sum_{cyc}{\frac{b+c}{a}}=92= \sum_{sym}{\frac{a^2b}{abc}}$ And $ \frac{AO}{OA'}\times \frac{BO}{OB'}\times \frac{AO}{OB'}=\frac{b+c}{a}\times \frac{c+a}{b} \times \frac{b+a}{c}=2+\sum_{sym}{\frac{a^2b}{abc}}=94$