Example 9 Given a rectangular cuboid $A B C D-A_{1} B_{1} C_{1} D_{1}$, where $A B=a, B C=b, C C_{1}=c$. Find the distance between the plane $A_{1} B D$ and the plane $B_{1} D_{1} C$.
Solution: Establish a spatial rectangular coordinate system as shown in Figure 11, then
$$
\begin{array}{l}
D(0,0,0) \\
A_{1}(b, 0, c) \\
B(b, a, 0) \text {, } \\
C(0, a, 0) .
\end{array}
$$

Thus, $D A_{1}=(b, 0, c), D B=(b, a, 0)$,
$$
D C=(0, a, 0) \text {. }
$$

Let the normal vector of plane $A_{1} B D$ be $\boldsymbol{n}=(x, y, z)$.
From $n \perp D A_{1}, n \perp D B$, we get
$$
b x+c z=0, b x+a y=0 .
$$

Let $x=a c$, then $y=-b c, z=-a b$.
Therefore, $n=(a c,-b c,-a b)$. To find the distance from point $C$ to plane $A_{1} B D$, we have
$$
d=\frac{|n \cdot D C|}{|n|}=\frac{|(a c,-b c,-a b) \cdot(0, a, 0)|}{\sqrt{a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}}} \text {. }
$$
$$
\frac{a b c}{\sqrt{a^{2} b^{2}+b^{2} c^{2}+a^{2} c^{2}}} \text {. }
$$