Three, (35 points) A group of students took an exam with 3 multiple-choice questions, each with four options. It is known that any two students in this group have at most one answer in common, and if one more student is added, regardless of their answers, the above property no longer holds. How many students are there in this group at minimum?
Three, this group of students must have at least 8 people.
First, consider if this group of students has 8 people, then it can meet the conditions of the problem.
Let the answers of these 8 people be $(1,1,1),(1,2,2),(2,2,1)$, $(2,1,2),(3,4,3),(3,3,4),(4,3,3),(4,4,4)$, which will suffice.

On the other hand, assume this group of students has only 7 people, then by the pigeonhole principle, there exists an $i, 1 \leqslant i \leqslant 4$, such that the number of people whose answer to the first question is $i$ is at most 1, let this person's answer be $(i, j, k)$. Thus, among the 16 different answers where the first question is $i$, it can only cover 7 of $(i, j, x)$ and $(i, x, k)$, the remaining 9 must be covered by other answer sets, which requires 9 more sets, leading to a contradiction. Therefore, this group of students must have at least 8 people.