Example 10 Given a unit cube $A B C D$ $A_{1} B_{1} C_{1} D_{1}$, $M$ and $N$ are the midpoints of $B B_{1}$ and $B_{1} C_{1}$, respectively, and $P$ is the midpoint of line segment $M N$. Find the distance between $D P$ and $A C_{1}$.
Solution: Establish a spatial rectangular coordinate system as shown in Figure 12, then
$$
\begin{array}{l}
B_{1}(0,0,0), \\
A(0,1,1), \\
C_{1}(1,0,0), \\
D(1,1,1), \\
P\left(\frac{1}{4}, 0, \frac{1}{4}\right)
\end{array}
$$

Let the equation of the plane $\alpha$ passing through $D P$ and parallel to $A C_{1}$ be
$$
A_{2} x+B_{2} y+C_{2} z+e=0 .
$$

Since $D P \in \alpha$, we have
$$
\begin{array}{l}
A_{2}+B_{2}+C_{2}+e=0, \\
\frac{1}{4} A_{2}+\frac{1}{4} C_{2}+e=0 .
\end{array}
$$

Also, since $\alpha \parallel A C_{1}$, the normal vector $\boldsymbol{n}$ of $\alpha$ is perpendicular to $\boldsymbol{A C}$. Thus, $A_{2}-B_{2}-C_{2}=0$.
From the above equations, we can find
$$
A_{2}=-\frac{1}{2} e, B_{2}=3 e, C_{2}=-\frac{7}{2} e,
$$

The equation of plane $\alpha$ is $x-6 y+7 z-2=0$.
Therefore, the distance from point $C_{1}$ to plane $\alpha$ is
$$
d=\frac{|1 \times 1-6 \times 0+7 \times 0-2|}{\sqrt{1^{2}+(-6)^{2}+7^{2}}}=\frac{\sqrt{86}}{86} .
$$