6. Given a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $6, M, N$ are points on $B B_{1}, B_{1} C_{1}$ respectively, $B_{1} M=B_{1} N=2, S, P$ are the midpoints of segments $A D, M N$ respectively, then the distance between the skew lines $S P$ and $A C_{1}$ is $\qquad$ .
As shown in the figure, with $\overrightarrow{D A}, \overrightarrow{D C}, \overrightarrow{D D_{1}}$ as the positive directions of the $x, y, z$ axes, respectively, we establish a rectangular coordinate system. Then, $A(6,0,0), C_{1}(0,6,6), S(3,0,0)$, and $P(5,6,5)$. Therefore, $\overrightarrow{A C_{1}}=(-6,6,6), \overrightarrow{S P}=(2,6,5)$. Let $\vec{n} \perp \overrightarrow{A C_{1}}, \vec{n} \perp \overrightarrow{S P}, \vec{n}=(x, y, z)$ $\Rightarrow\left\{\begin{array}{l}-x+y+z=0, \\ 2 x+6 y+5 z=0 .\end{array}\right.$ Taking $\vec{n}=(1,-7,8)$, the distance $d$ between the skew lines $S P$ and $A C_{1}$ is $d=\left|\overrightarrow{S A} \cdot \overrightarrow{n_{0}}\right|=\frac{3}{\sqrt{114}}=\frac{\sqrt{114}}{38}$