Three, (50 points) Let $A=\{1,2, \cdots, 30\}$. Find the smallest positive integer $n$, such that for any 11 subsets of $A$, if the union of any 5 of them has at least $n$ elements, then there must exist 3 of these 11 subsets whose intersection is non-empty.
Three, the minimum value of $n$ is 22.
First, prove: $n \geqslant 22$.
$$
\begin{array}{l}
\quad \text { Let } A_{i}=\{i, i+5, i+10, i+15, i+20, i+25\}, \\
i=1,2, \cdots, 5, \\
\quad B_{j}=\{j, j+6, j+12, j+18, j+24\}, j= \\
1,2, \cdots, 6 .
\end{array}
$$
$$
\begin{array}{l}
\text { Clearly, }\left|A_{i}\right|=6(i=1,2, \cdots, 5), \\
\left|A_{i} \cap A_{j}\right|=0(1 \leqslant i<j \leqslant 5), \\
\left|B_{j}\right|=5(j=1,2, \cdots, 6), \\
\left|B_{i} \cap B_{j}\right|=0(1 \leqslant i<j \leqslant 6), \\
\left|A_{i} \cap B_{j}\right|=1(1 \leqslant i \leqslant 5,1 \leqslant j \leqslant 6) .
\end{array}
$$

Thus, for any three subsets, there must be two that are both $A_{i}$ or both $B_{j}$, and their intersection is the empty set. For any five subsets $A_{i_{1}}, A_{i_{2}}, \cdots, A_{i_{s}}, B_{j_{1}}, B_{j_{2}}, \cdots$, $B_{j_{t}}(s+t=5)$, we have
$$
\begin{aligned}
& \left|A_{i_{1}} \cup A_{i_{2}} \cup \cdots \cup A_{i_{s}} \cup B_{j_{1}} \cup B_{j_{2}} \cup \cdots \cup B_{j_{t}}\right| \\
= & \left|A_{i_{1}}\right|+\left|A_{i_{2}}\right|+\cdots+\left|A_{i_{s}}\right|+\left|B_{j_{1}}\right|+ \\
& \left|B_{j_{2}}\right|+\cdots+\left|B_{j_{t}}\right|-s t \\
= & 6 s+5 t-s t=6 s+5(5-s)-s(5-s) \\
= & s^{2}-4 s+25 \geqslant 21,
\end{aligned}
$$

and the intersection of any three subsets is the empty set, hence $n \geqslant 22$.
Next, prove: $n=22$ satisfies the conditions.
Assume there exist 11 subsets of $A$, such that the union of any five of them contains at least 22 elements, and the intersection of any three is the empty set. Since each element belongs to at most two subsets, we can assume each element belongs to exactly two subsets (otherwise, add some elements to some subsets, and the conditions still hold). By the pigeonhole principle, among these 11 subsets, there must be one subset (let it be $B$) that contains at least $\left[\frac{2 \times 30}{11}\right]+1=6$ elements. Let the other 10 subsets be $B_{1}, B_{2}, \cdots, B_{10}$, then, any five subsets not containing $B$ correspond to at least 22 elements in $A$, and all 5-subset groups not containing $B$ together correspond to at least $22 \mathrm{C}_{10}^{5}$ elements.

On the other hand, for an element $a$, if $a \notin B$, then two of $B_{1}, B_{2}, \cdots, B_{10}$ contain $a$, so $a$ is counted $\mathrm{C}_{10}^{5}-\mathrm{C}_{8}^{5}$ times; if $a \in B$, then one of $B_{1}, B_{2}, \cdots, B_{10}$ contains $a$, so $a$ is counted $\mathrm{C}_{10}^{5}-\mathrm{C}_{9}^{5}$ times. Thus,
$$
\begin{array}{l}
22 \mathrm{C}_{10}^{5} \leqslant(30-|B|)\left(\mathrm{C}_{10}^{5}-\mathrm{C}_{8}^{5}\right)+|B|\left(\mathrm{C}_{10}^{5}-\mathrm{C}_{9}^{5}\right) \\
=30\left(\mathrm{C}_{10}^{5}-\mathrm{C}_{8}^{5}\right)-|B|\left(\mathrm{C}_{9}^{5}-\mathrm{C}_{8}^{5}\right) \\
\leqslant 30\left(\mathrm{C}_{10}^{5}-\mathrm{C}_{8}^{5}\right)-6\left(\mathrm{C}_{9}^{5}-\mathrm{C}_{8}^{5}\right) .
\end{array}
$$

This leads to $25 \leqslant 24$, a contradiction.
In conclusion, the minimum value of $n$ is 22.