Let $L$ be a set of $n$ axis-parallel lines in $\mathbb{R}^3$. We are are interested in partitions of $\mathbb{R}^3$ by a set $H$ of three planes such that each open cell in the arrangement $\mathcal{A}(H)$ is intersected by as few lines from $L$ as possible. We study such partitions in three settings, depending on the type of splitting planes that we allow. We obtain the following results. $\bullet$ There are sets $L$ of $n$ axis-parallel lines such that, for any set $H$ of three splitting planes, there is an open cell in $\mathcal{A}(H)$ that intersects at least~$\lfloor n/3 \rfloor-1 \approx \frac{1}{3}n$ lines. $\bullet$ If we require the splitting planes to be axis-parallel, then there are sets $L$ of $n$ axis-parallel lines such that, for any set $H$ of three splitting planes, there is an open cell in $\mathcal{A}(H)$ that intersects at least $\frac{3}{2}\lfloor n/4 \rfloor -1 \approx \left( \frac{1}{3}+\frac{1}{24}\right) n$ lines. Furthermore, for any set $L$ of $n$ axis-parallel lines, there exists a set $H$ of three axis-parallel splitting planes such that each open cell in $\mathcal{A}(H)$ intersects at most $\frac{7}{18} n = \left( \frac{1}{3}+\frac{1}{18}\right) n$ lines. $\bullet$ For any set $L$ of $n$ axis-parallel lines, there exists a set $H$ of three axis-parallel and mutually orthogonal splitting planes, such that each open cell in $\mathcal{A}(H)$ intersects at most $\lceil \frac{5}{12} n \rceil \approx \left( \frac{1}{3}+\frac{1}{12}\right) n$ lines.

翻译：设$L$为$\mathbb{R}^3$中$n$条轴平行直线的集合。我们关注由三个平面构成的集合$H$对$\mathbb{R}^3$的划分，使得在排列$\mathcal{A}(H)$中每个开单元被$L$中尽可能少的直线所相交。根据所允许的切割平面类型，我们在三种设定下研究此类划分，并获得以下结果。 $\bullet$ 存在由$n$条轴平行直线构成的集合$L$，使得对于任意三个切割平面构成的集合$H$，在$\mathcal{A}(H)$中至少有一个开单元与至少$\lfloor n/3 \rfloor-1 \approx \frac{1}{3}n$条直线相交。 $\bullet$ 若要求切割平面为轴平行，则存在由$n$条轴平行直线构成的集合$L$，使得对于任意三个轴平行切割平面构成的集合$H$，在$\mathcal{A}(H)$中至少有一个开单元与至少$\frac{3}{2}\lfloor n/4 \rfloor -1 \approx \left( \frac{1}{3}+\frac{1}{24}\right) n$条直线相交。此外，对于任意由$n$条轴平行直线构成的集合$L$，存在三个轴平行切割平面构成的集合$H$，使得$\mathcal{A}(H)$中每个开单元至多与$\frac{7}{18} n = \left( \frac{1}{3}+\frac{1}{18}\right) n$条直线相交。 $\bullet$ 对于任意由$n$条轴平行直线构成的集合$L$，存在三个轴平行且两两正交的切割平面构成的集合$H$，使得$\mathcal{A}(H)$中每个开单元至多与$\lceil \frac{5}{12} n \rceil \approx \left( \frac{1}{3}+\frac{1}{12}\right) n$条直线相交。