Three, (50 points) A conference was attended by $12 k$ people $(k \in$ $\mathbf{N}$ ), where each person has greeted exactly $3 k+6$ other people. For any two people, the number of people who have greeted both of them is the same. How many people attended the conference?
Three, connecting the points of people and their acquaintances, each point forms $\mathrm{C}_{3 k+6}^{2}$ angles, each simple connection corresponds to a vertex, so the graph has a total of $12 k \cdot C_{3 k+6}^{2}$ angles. On the other hand, suppose for any two people, there are $n$ people who know both of them, thus for each pair, there are $n$ angles, each angle uniquely corresponding to a pair of people, and for each pair, there are exactly $n$ angles, from this perspective, the total number of angles is $n \cdot \mathrm{C}_{12 k}^{2}$.

Of course, $n \cdot \mathrm{C}_{12 k}^{2}=12 k \cdot \mathrm{C}_{3 k+6}^{2}$, simplifying gives $9 k^{2}+33 k -12 n k+n+30=0$, where $n, k$ are integers. It is easy to see that $3 \mid n$, let $n=3 m$, further simplifying gives $3 k^{2}+11 k-12 k m+m+10=0$, i.e., $4 m=\frac{k+3+(9 k+43)}{12 k-1}$. When $k \geqslant 15$, $12 k -1>9 k+43, 4 m$ cannot be an integer. For $1 \leqslant k \leqslant 14$, only when $k=3$, $\frac{9 k+43}{12 k-1}$ is an integer. Therefore, it can only be 36 people.
Below is a construction to show that the proposition holds.
\begin{tabular}{|c|c|c|c|c|c|}
\hline$R$ & $O$ & $Y$ & $G$ & $B$ & $V$ \\
\hline$V$ & $R$ & $O$ & $Y$ & $G$ & $B$ \\
\hline$B$ & $V$ & $R$ & $O$ & $Y$ & $G$ \\
\hline$G$ & $B$ & $V$ & $R$ & $O$ & $Y$ \\
\hline$Y$ & $G$ & $B$ & $V$ & $R$ & $O$ \\
\hline$O$ & $Y$ & $G$ & $B$ & $V$ & $R$ \\
\hline
\end{tabular}

In the table, $R, O, Y, G, B, V$ represent 6 colors. Color the people according to the above arrangement, each person knows exactly the people sitting in the same row, the same column, or having the same color as them, thus ensuring that each person knows exactly 15 people. It only remains to prove that this construction satisfies the second condition.

Let $P, Q$ be any two attendees. If they are in the same row, the people who know both of them are the other 4 people in this row, plus the person in the same column as $P$ and the same color as $Q$, and the person in the same column as $Q$ and the same color as $P$. If $P, Q$ are of the same color or in the same column, the analysis is similar. Assuming $P$ and $Q$ are not in the same row, not in the same column, and have different colors, the 6 people they both know are: the person in $P$'s row and $Q$'s column (different in color from both $P$ and $Q$), the person in $P$'s column and $Q$'s row, the 2 people in $P$'s row and column and the same color as $Q$, and the 2 people in $Q$'s row and column and the same color as $P$.