Example 3 As shown in Figure 1.13.3, given $\frac{A P}{P D}+\frac{B P}{P E}+\frac{C P}{P F}=90$, find the value of $\frac{A P}{P D} \cdot \frac{B P}{P E} \cdot \frac{C P}{P F}$.
Let $\frac{A P}{P D}=x, \frac{B P}{P E}=y, \frac{C P}{P F}=z, P$ be the center of mass of the point group $A, B, C$, and $\frac{B D}{D C}=a$. Since $P$ is the center of mass of points $A, D$, we take $m_{A}=1, m_{D}=x$. Also, since $D$ is the center of mass of points $B, C$, we have $m_{B}=\frac{x}{1+a}$, $m_{C}=\frac{a x}{1+a}$. Furthermore, since $P$ is the center of mass of points $C, F$, we get $m_{F}=\frac{a x z}{1+a}$.
Similarly, $m_{E}=\frac{\gamma x}{1+a}$. According to property 3, $F$ and $E$ are the centers of mass of the point groups $A, B$ and $A, C$, respectively, so
$$
1+\frac{x}{1+a}=\frac{a x z}{1+a}, 1+\frac{a x}{1+a}=\frac{y x}{1+a}
$$

Solving for $y, z$, we find
$$
\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=\frac{1}{x+1}+\frac{x}{1+a+a x+x}+\frac{x a}{1+a+a x+x}=1
$$

Thus,
$$
x y z=(x+y+z)+2=94
$$