6.52 In an exam of a class, there are 3 multiple-choice questions, each with 4 options. It is known that a group of students' answers have property $P$: any two students' answers have at most 1 question with the same answer. Moreover, if this group of students is increased by 1 person, regardless of this person's answers, property $P$ no longer holds. How many students are there in this group at minimum?
[Solution] Number the 4 options for each question as $0,1,2,3$. Each person's answer can be written as a triplet $(i, j, k), i, j, k=0,1,2,3$. Thus, all 64 possible answer sets correspond to the 64 nodes (including vertices) of a $3 \times 3 \times 3$ cube divided into 27 unit cubes. The 4 planes of the cube with nodes (differing by the 3rd component) are drawn separately, and the selected nodes are marked as follows:
It is easy to see that no two black points lie on the same grid line, but for any other node, there is a black point on the same grid line. This indicates that when there are 8 students, and their answers are $(0,0,0),(1,1,0),(1,0,1),(0,1,1),(3,2,2),(2,3,2),(2,2,3),(3,3,3)$, all the requirements of the problem are satisfied.

On the other hand, for 7 students, their answer arrays correspond to 7 nodes of the cube, referred to as black points. By the pigeonhole principle, among the 4 planes shown, there is always one plane with at most 1 black point. On this plane, excluding the two grid lines where the black point lies (if any), there are at least 9 nodes. On the other planes, there are at most 7 black points, and each black point's grid line can contain at most 1 of the 9 points. Therefore, it is still possible to select 1 point from the 9 points, such that when the 8th student, represented by this point, joins the first 7, property $P$ still holds. This means that no group of 7 students can meet all the requirements of the problem.
In summary, the minimum number of students required to meet all the requirements of the problem is 8.