jmc

Let $K_A=[BOC],K_B=[COA],$ and $K_C=[AOB].$ Due to triangles $BOC$ and $ABC$ having the same base,$$\frac{AO}{O{A}^{\prime }}+1=\frac{A{A}^{\prime }}{O{A}^{\prime }}=\frac{[ABC]}{[BOC]}=\frac{K_A+K_B+K_C}{K_A}.$$Therefore, we have$$\frac{AO}{O{A}^{\prime }}=\frac{K_B+K_C}{K_A}$$$$\frac{BO}{O{B}^{\prime }}=\frac{K_A+K_C}{K_B}$$$$\frac{CO}{O{C}^{\prime }}=\frac{K_A+K_B}{K_C}.$$Thus, we are given$$\frac{K_B+K_C}{K_A}+\frac{K_A+K_C}{K_B}+\frac{K_A+K_B}{K_C}=92.$$Combining and expanding gives$$\frac{K_A^2{K}_B+K_A{K}_B^2+K_A^2{K}_C+K_A{K}_C^2+K_B^2{K}_C+K_B{K}_C^2}{K_A{K}_B{K}_C}=92.$$We desire $\frac{(K_B+K_C)(K_C+K_A)(K_A+K_B)}{K_A{K}_B{K}_C}.$ Expanding this gives$$\frac{K_A^2{K}_B+K_A{K}_B^2+K_A^2{K}_C+K_A{K}_C^2+K_B^2{K}_C+K_B{K}_C^2}{K_A{K}_B{K}_C}+2=\boxed{94}.$$