87. Let $A=a c e+b d f, B=a b c+b c d+c d e+d e f+e f a+f a b$, then by the AM-GM inequality we have
$$\begin{array}{c} A+B=(a+d)(b+e)(c+f) \leqslant\left[\frac{(a+d)+(b+e)+(c+f)}{3}\right]^{3}=\frac{1}{27} \\ B \leqslant \frac{1}{27}-A \leqslant \frac{1}{27}-\frac{1}{108}=\frac{1}{36} \end{array}$$

Equality holds if and only if $a+d=b+e=c+f=\frac{1}{3}$, and $a c e+b d f=\frac{1}{108}$.
That is, when $a c e+\left(\frac{1}{3}-a\right)\left(\frac{1}{3}-c\right)\left(\frac{1}{3}-e\right)=\frac{1}{108}$ and $0<a, c, e<\frac{1}{3}$. At this time, $a c+ c e+e a+\frac{1}{12}=\frac{1}{3}(a+c+e)$

For example, when $a=\frac{1}{4}, b=\frac{1}{5}, c=\frac{2}{9}, d=\frac{1}{12}, e=\frac{2}{15}, f=\frac{1}{9}$, equality holds.