$5 \cdot 120$ A certain meeting was attended by $12 k$ people, each of whom greeted exactly $3 k+6$ others. For any two people, the number of people who greeted both of them is the same. How many people attended the meeting?
[Solution] Let the number of people who have greeted any two individuals be $n$.

Fix any person $a$, and let the set of people who have greeted $a$ be denoted as $B$, and the set of people who have not greeted $a$ be denoted as $C$. Thus, $|B|=3k+6, |C|=9k-7$.

Take any $b \in B$, the people who have greeted both $a$ and $b$ must all be in $B$. Therefore, $b$ has greeted $n$ people in $B$. $b$ has greeted $3k+5-n$ people in $C$. Now take any $c \in C$, the people who have greeted both $a$ and $c$ must all be in $B$. Thus, $c$ has greeted $n$ people in $B$. The number of greeting pairs between $B$ and $C$ is
$$
\begin{array}{c}
(3k+6)(3k+5-n)=(9k-7)n. \\
9k^2-12kn+33k+n+30=0.
\end{array}
$$

Since all terms in the above equation except $n$ are multiples of 3, it must be that $n=3m, m \in \mathbb{N}$. This gives us
$$
4m=k+3+\frac{9k+43}{12k-1}.
$$

When $k \geq 15$, $9k+43 < 12k-1$, so $12k-1$ can only be 25, 35, or 175. Clearly, only 35 is feasible. In this case, $k=3, m=2, n=6, 3k+6=15$. The only possible scenario that meets the requirements is that each person has greeted 15 people, and the number of people who have greeted any two people is 6.
We now provide an example to show that the above scenario is feasible.

Suppose there are 36 people attending the meeting, with 6 people each wearing shirts of 6 different colors: red, orange, yellow, green, blue, and purple. We denote the people wearing these 6 colors as $R$, $O$, $Y$, $G$, $B$, and $V$. Then, let these 36 people sit in 6 rows and 6 columns, marked by their shirt colors as shown in the table below. It is stipulated that each person greets exactly those in the same row, same column, and same color. Thus, each person greets exactly 15 people. Therefore, we only need to verify that the number of people who have greeted any two people is exactly 6.
(1) When $P$ and $Q$ are in the same row, the other 4 people in the same row, the person in the same column as $P$ and the same color as $Q$, and the person in the same color as $P$ and the same column as $Q$ are the 6 people who have greeted both $P$ and $Q$. All others have greeted at most one of $P$ or $Q$.
(2) When $P$ and $Q$ are in the same column or the same color, similarly, the number of people who have greeted both $P$ and $Q$ is exactly 6.
(3) Suppose $P$ and $Q$ are neither in the same row, nor in the same column, nor of the same color. In this case, the person in the same row as $P$ and the same column as $Q$, the person in the same column as $P$ and the same row as $Q$, the person in the same row as $P$ and the same color as $Q$, the person in the same color as $P$ and the same row as $Q$, the person in the same column as $P$ and the same color as $Q$, and the person in the same color as $P$ and the same column as $Q$ are the 6 people who have greeted both $P$ and $Q$.
In summary, there are 36 people attending the meeting.