12 * $2 \star$ As shown in Figure a, given that the edge length of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ is 2, and point $E$ is the midpoint of edge $C D$. Find the distance between the skew lines $A_{1} C_{1}$ and $B_{1} E$.
Parse as shown in Figure b, establish the spatial Cartesian coordinate system $D-xyz$, then $A_{1}(2,0,2), C_{1}(0,2,2)$, $B_{1}(2,2,2), E(0,1,0), \overrightarrow{A_{1} C_{1}}=(-2,2,0), \overrightarrow{B_{1} E}=(-2,-1,-2), \overrightarrow{A_{1} B_{1}}=$ $(0,2,0)$.

Let $P Q$ be the common perpendicular segment of the skew lines $A_{1} C_{1}$ and $B_{1} E$, and $\overrightarrow{A_{1} P}=\lambda \overrightarrow{A_{1} C_{1}}=(-2 \lambda, 2 \lambda, 0)$, and $\mathbf{a}$
Group b
$\overrightarrow{B_{1} Q}=\mu \overrightarrow{B_{1} E}=(-2 \mu,-\mu,-2 \mu)$. Then $\overrightarrow{P Q}=\overrightarrow{P A_{1}}+\overrightarrow{A_{1} B_{1}}+\overrightarrow{B_{1} Q}=(2 \lambda-2 \mu$, $-2 \lambda-\mu+2,-2 \mu)$.

Since $\overrightarrow{P Q} \perp \overrightarrow{A_{1} C_{1}}, \overrightarrow{P Q} \perp \overrightarrow{B_{1} E}$, we have $\overrightarrow{P Q} \cdot \overrightarrow{A_{1} C_{1}}=0, \overrightarrow{P Q} \cdot \overrightarrow{B_{1} E}=0$. Also, $\overrightarrow{P Q} \cdot \overrightarrow{A_{1} C_{1}}=-8 \lambda+2 \mu+4, \overrightarrow{P Q} \cdot \overrightarrow{B_{1} E}=-2 \lambda+9 \mu-2$, then $\left\{\begin{array}{l}-8 \lambda+2 \mu+4=0, \\ -2 \lambda+9 \mu-2=0 .\end{array}\right.$ Solving these, we get $\lambda=$ $\frac{10}{17}, \mu=\frac{6}{17}$. Therefore, $\overrightarrow{P Q}=\left(\frac{8}{17}, \frac{8}{17},-\frac{12}{17}\right)$,
$$
|\overrightarrow{P Q}|=\sqrt{\left(\frac{8}{17}\right)^{2}+\left(\frac{8}{17}\right)^{2}+\left(-\frac{12}{17}\right)^{2}}=\frac{4 \sqrt{17}}{17}
$$