Three, let $S=\{1,2, \cdots, 15\}$. From $S$, take $n$ subsets $A_{1}, A_{2}, \cdots, A_{n}$ that satisfy the following conditions:
(i) $\left|A_{i}\right|=7, i=1,2, \cdots, n$;
(ii) $\left|A_{i} \cap A_{j}\right| \leqslant 3,1 \leqslant i<j \leqslant n$;
(iii) For any three-element subset $M$ of $S$, there exists some $A_{k}$ such that $M \subset A_{k}$.
Find the minimum value of the number of such subsets $n$.
Let $\mathscr{A}=\left\{A_{1}, A_{2}, \cdots, A_{n}\right\}$ be any family of sets that satisfies the conditions of the problem. For any $a \in S$, let the number of sets in the family $\mathscr{A}$ that contain $a$ be denoted by $r(a)$. These $r(a)$ sets each contain $C_{6}^{2}:=15$ three-element subsets that include $a$. On the other hand, the three-element subsets of $S$ are contained in the sets of the family, so,
$$
15 r(a) \geqslant 91, \quad r(a) \geqslant 7.
$$

By calculating the total number of elements in the sets of the family $\mathscr{A}$ in two ways, we see that
$$
n \times 7=\sum_{a \in S} r(a) \geqslant 15 \times 7, \quad n \geqslant 15.
$$

Next, we construct a specific family of sets $S=\left\{A_{1}, A_{2}, \cdots, A_{15}\right\}$ that satisfies the conditions of the problem. First, we label the 15 elements of $S=\{1,2, \cdots, 15\}$ in a clockwise order on the 15 equally spaced points on a circle (as shown in Figure 1).

We define $A_{1}=\{1,2,4,5,6,11,13\}$, and let $A_{j}$ (for $j=2,3, \cdots, 15$) be the 7-element subset of $S$ obtained by rotating $A_{1}$ clockwise by $j-1$ arc segments. Here, an arc segment is the arc between two adjacent points on the circle divided into 15 equal parts. Figure 2 shows the diagram of $A_{1}$. The numbers marked on the inner side of the arc segments in the figure represent the number of arc segments contained in each arc segment.

We now verify that such a $\mathscr{A}=\left\{A_{1}, A_{2}, \cdots, A_{15}\right\}$ satisfies the conditions of the problem.
(i) Clearly, each set in $\mathscr{A}$ consists of exactly 7 elements of $S$.
(ii) If $1 \leqslant j-i \leqslant 7$, then rotating $A_{i}$ clockwise by $j-i$ arc segments results in $A_{j}$. Analyzing the diagram of $A_{1}$ (Figure 2), we see that there are exactly three pairs of elements for each of the distances
$$
1,2,3,4,5,6,7
$$

in the clockwise direction. Each $A_{i}$ has the same property. Rotating $A_{i}$ clockwise by $j-i$ arc segments $(1 \leqslant j-i \leqslant 7)$, exactly three elements of $A_{i}$ will move to positions where the elements are still in $A_{i}$. Therefore,
$$
\left|A_{i} \cap A_{j}\right|=3.
$$

If $8 \leqslant j-i \leqslant 14$, then rotating $A_{j}$ clockwise by $15-(j-i)$ arc segments results in $A_{i}$. Similarly, we can conclude that
$$
\left|A_{i} \cap A_{j}\right|=3.
$$
(iii) Let $M=\{u, v, w\}$ be any three-element subset of $S$. Without loss of generality, assume $u, v, w$ are arranged in a clockwise order on the circle. Consider the number of arc segments between $u$ and $v$, $v$ and $w$, and $w$ and $u$ in the clockwise direction, and denote the pair of the smaller two numbers of these arc segments as $(a, b)$ (in clockwise order). We will classify the three-element subsets of $S$ based on these pairs. Clearly,
$$
1 \leqslant a, b \leqslant 7,
$$

and except for the pair $(5,5)$, all other pairs $(a, b)$ satisfy
$$
a+b \leqslant 9.
$$
Each pair $(a, b)$ type appears exactly once in $A_{1}$:
$$
\begin{array}{l}
(1,1),(1,2),(2,1),(1,3),(3,1),(1,4),(4,1), \\
(1,5),(5,1),(1,6),(6,1),(1,7),(7,1) ; \\
(2,2),(2,3),(3,2),(2,4),(4,2),(2,5),(5,2), \\
(2,6),(6,2),(2,7),(7,2) ; \\
(3,3),(3,4),(4,3),(3,5),(5,3),(3,6,),(6,3) ; \\
(4,4),(4,5),(5,4) ; \\
(5,5) ;
\end{array}
$$

Accordingly, any three-element subset $M$ of $S$ is contained in at least one set of the family $\mathscr{A}$ (in fact, exactly one set of the family $\mathscr{A}$).