3

3 committies of 9 people equals 27 people in total ( 3 X 9 = 27)
9 devided by 3 committies = 3 minimum.

The minimum number is 1.

Committee 1 leaves 4 people in no committee.
Committee 2 is most spread if those 4 people are in it, meaning 5 people are in both 1 & 2.
Committee 3 can have the 4 people who aren’t in committee 1, and the 4 people who aren’t in committee 2, leaving 1 person who is in both, who ends up needing to be in committee 3 too, like this:

Committee 1: 1 1 1 1 1 1 1 1 1 0 0 0 0
Committee 2: 0 0 0 0 2 2 2 2 2 2 2 2 2
Committee 3: 3 3 3 3 3 0 0 0 0 3 3 3 3

That is correct. Here is how I thought of it. Committee 2 can have at most 4 people who are not in committee 1, meaning committee 1 and committee 2 have 5 people in common, leaving 8 who are not in committee 1 and committee 2. Committee 3 can have all those 8 people, leaving one more who must be in both committee 1 and committee 2.

Here is the Math Stack Exchange way of doing it. Instead of looking at people who are in committees, look at people who are not in at least one committee. We then subtract that number from 13 to get the number who must be in all 3 committees.

Committee 1 has 4 people who are not in it. Let’s designate them as 1,2,3 and 4. Committee 2 also has 4 people not in it. To maximize the total, we take people not already excluded from committee 1. Let’s designate them as 5,6,7 and 8. Similarly, we have people 9,10, 11 and 12 excluded from committee 3. We have excluded people 1 to 12 from at least one committee, the maximum possible. The remaining person, 13, must be in all 3 committees.
This method generalizes easily.

Yes. I think your reasoning, my reasoning, and Math Stack Exchange, are all the same reasoning, just expressed in different words.

I divided 13 into three groups of 4/4/5.
Then I assigned two groups to each committee.
The committee that was assigned the two groups of 4 would need a ninth member.
Therefore the answer is 1.