  We prove that if a family of compact connected sets in the plane has the
property that every three members of it are intersected by a line, then there
are three lines intersecting all the sets in the family. This answers a
question of Eckhoff from 1993, who proved that, under the same condition, there
are four lines intersecting all the sets. In fact, we prove a colorful version
of this result, under weakened conditions on the sets.
  A triple of sets $A,B,C$ in the plane is said to be a {\em tight} if
$\textrm{conv}(A\cup B)\cap \textrm{conv}(A\cup C)\cap \textrm{conv}(B\cap
C)\neq \emptyset.$ This notion was first introduced by Holmsen, where he showed
that if $\mathcal{F}$ is a family of compact convex sets in the plane in which
every three sets form a tight triple, then there is a line intersecting at
least $\frac{1}{8}|\mathcal{F}|$ members of $\mathcal{F}$. Here we prove that
if $\mathcal{F}_1,\dots,\mathcal{F}_6$ are families of compact connected sets
in the plane such that every three sets, chosen from three distinct families
$\mathcal{F}_i$, form a tight triple, then there exists $1\le j\le 6$ and three
lines intersecting every member of $\mathcal{F}_j$. In particular, this
improves $\frac{1}{8}$ to $\frac{1}{3}$ in Holmsen's result.
