We want to color the three-element parts of $\{1,2,3,4,5,6,7\}$, such that if two of these parts have no element in common, then they must be of different colors. What is the minimum number of colors needed to achieve this goal?
Consider the sequence of sets $\{1,2,3\},\{4,5,6\},\{1,2,7\},\{3,4,6\},\{1,5,7\},\{2,3,6\},\{4,5,7\}$, $\{1,2,3\}$.
Each set must have a different color from the next, so there must be at least two colors. If there were exactly two colors, then the colors would have to alternate, which is impossible because the last set is the same as the first and should be of the opposite color.
Conversely, let's show that three colors suffice:

- We color blue the sets that contain at least two elements among 1, 2, 3.
- We color green the sets not colored blue and that contain at least two elements among 4, 5, 6.
- We color red the sets not colored blue or green.

It is clear that two blue sets have an element in common among 1, 2, 3; similarly, two green sets have an element in common among 4, 5, 6. Finally, any red set contains three elements, which necessarily includes the element 7: two red sets therefore also have an element in common.