Example 14 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 $A C_{1}$.
(1982 Shanghai High School Competition Problem)
To establish a rectangular coordinate system for figure 20-13, we have
$$
B_{1}(0,0,0), A(0,1,1), C_{1}(1,0,0), D(1,1,1), P\left(\frac{1}{4}, 0, \frac{1}{4}\right).
$$

Let the equation of the plane $\alpha$ passing through $DP$ and parallel to $AC_{1}$ be $A_{2} x+B_{2} y+C_{2} z+e=0$.
Since $DP \in \alpha$, we have $A_{2}+B_{2}+C_{2}+e=0, \frac{1}{4} A_{2}+\frac{1}{4} C_{2}+e=0$.
Also, since $\alpha \parallel AC_{1}$, then $\overrightarrow{AC_{1}} \perp \alpha$'s normal vector $\boldsymbol{n}$.
Thus, $A_{2}-B_{2}-C_{2}=0$.
From the above equations, we can solve for $A_{2}=-\frac{1}{2} e, B_{2}=3 e, C_{2}=-\frac{7}{2} e$,
Therefore, the equation of the plane $\alpha$ is $x-6 y+7 z-2=0$.
Hence, the distance from point $C_{1}$ to the plane $\alpha$ is $d=\frac{|1 \cdot 1-6 \cdot 0+7 \cdot 0-2|}{\sqrt{1^{2}+\left(-6^{2}\right)+7^{2}}}=\frac{\sqrt{86}}{86}$.