5. (Ireland) A conference is attended by $12 k$ people, each of whom has greeted exactly $3 k+6$ others. For any two people, the number of people who have greeted both of them is the same. How many people attended the conference?
For any two people, let the number of other people who have greeted these two people be $n$. By the problem, $n$ is fixed. For a specific person $a$, let $\mathrm{B}$ be the set of all people who have greeted $a$, and $\mathrm{C}$ be the set of people who have not greeted $a$. Then, $\mathbf{B}$ contains $3 k+6$ people, and $\mathrm{C}$ contains $9 k-7$ people. For any person $b$ in $\mathbf{B}$, the people who have greeted both $a$ and $b$ must be in $\mathbf{B}$. Therefore, $b$ has greeted $n$ people in $\mathbf{B}$ and $3 k+5-n$ people in $\mathrm{C}$. For any person $c$ in $\mathrm{C}$, the people who have greeted both $a$ and $c$ must also be in $\mathbf{B}$. Thus, $c$ has greeted $n$ people in $\mathbf{B}$. The total number of greetings between $\mathbf{B}$ and $\mathbf{C}$ is 
$(3 k+6)(3 k+5-n)=(9 k-7) n$.
Simplifying, we get $9 k^{2}-12 k n+33 k+n+30=0$. From this, we know that $n=3 m$, where $m$ is a positive integer, and $4 m=k+3+\frac{9 k+43}{12 k-1}$. When $k \geqslant 15$, $12 k-1>9 k+43$, and $4 m$ is not an integer. For $1 \leqslant k \leqslant 14$, only when $k=3$, $\frac{9 k+43}{12 k-1}$ is an integer. Therefore, the only possibility is that 36 people attended the meeting. The following construction shows that the solution to the original problem indeed exists.
\begin{tabular}{ll} 
ROYGBV & $R:$ Red \\
VROYGB & $O:$ Orange \\
BVROYG & $Y$ : Yellow \\
GBVROY & $G$ : Green \\
YGBVRO & $B$ : Blue \\
OYGBVR & $V$ : Violet
\end{tabular}
People of the same color. Clearly, each person knows exactly 15 people. 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 that row, and the person in the column of $P$ with the color of $Q$, and the person in the column of $Q$ with the color of $P$. When $P, Q$ are in the same column or have the same color, similar analysis applies. Assume $P, Q$ are in different rows, different columns, and have different colors. Then the 6 people who know both of them are: the person in the row of $P$ and the column of $Q$, the person in the row of $P$ with the color of $Q$, the person in the column of $P$ and the row of $Q$, the person in the column of $P$ with the color of $Q$, the person with the color of $P$ in the row of $Q$, or the column of $Q$.