6. Let $X$ be a set of $n$ points in the plane, and for each point $A$ in $X$, there are exactly 3 points in $X$ that are at a distance of 1 from $A$. Then the minimum value of $n$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7
6.C.

A line segment of length 1 is called a "good edge".
For any two points in $X$, if their distance is 1, then they are connected by a line segment. Thus, each point gives rise to 3 good edges. Therefore, $n$ points give rise to $3n$ good edges in total. Since each good edge has two endpoints and is counted twice, the number of good edges is $\frac{3n}{2}$. Hence, $n$ must be even.

If $n=4$, then the distance between any two points in $X$ is 1. For any three points $A, B, C$, $\triangle ABC$ is an equilateral triangle. Similarly, $\triangle ABD$ is an equilateral triangle, which implies $CD > 1$, a contradiction. Therefore,
$$
n \geqslant 6 \text{.}
$$

When $n=6$, as shown in Figure 4, there exists a set of 6 points that meet the conditions.