Example 11 In the right quadrilateral prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, the base is a right trapezoid $A B C D, C D \perp A D, A B=$ $2, A D=3, D C=6, A A_{1}=6, M, N$ are the midpoints of $C_{1} D_{1}, C C_{1}$ respectively. Find the distance from $M N$ to the plane $A D_{1} C$.
Solution: Establish a rectangular coordinate system as shown in Figure 13, then
$$
\begin{array}{l}
D(0,0,0), \\
A(3,0,0), \\
C(0,6,0), \\
D_{1}(0,0,6), (0,6,3).
\end{array}
$$

Thus, $A C=(-3,6,0), A D_{1}=(-3,0,6)$.
Let the normal vector of plane $A D_{1} C$ be $\boldsymbol{n}=(x, y, z)$.
Since $\boldsymbol{n} \perp \boldsymbol{A C}, \boldsymbol{n} \perp \mathrm{AD}$ 1, we get
$$
-3 x+6 y=0, -3 x+6 z=0.
$$

Let $x=2$, then $y=z=1$.
Therefore, $\boldsymbol{n}=(2,1,1), N C=(0,0,-3)$.
To find the distance between $M N$ and plane $A D_{1} C$, we only need to find the distance from point $N$ to plane $A D_{1} C$, then
$$
d=\frac{|n \cdot N C|}{|n|}=\frac{3}{\sqrt{6}}=\frac{\sqrt{6}}{2}.
$$

Thus, the distance between $M N$ and plane $A D_{1} C$ is $\frac{\sqrt{6}}{2}$.
Finally, it should be noted that there are generally two methods to determine the normal vector of a plane:
(1) The method of undetermined coefficients. Set the normal vector $\boldsymbol{n}=(x, y, z)$, and determine the values of $x, y, z$ based on the perpendicular relationship between vectors.
(2) Find the equation of the plane $A x+B y+C z+D=0$, then the normal vector of the plane is $(A, B, C)$ (see Example 10).