Hkg TST

Solution:

No. Suppose such a choice exists. Let $n_1, n_2, \dots, n_{2019}$ be the number of points on the planes respectively. Note that $n_j \ge 3$ for $1 \le j \le 2019$. It is clear that
$$2024 = \binom{24}{3} = \binom{n_1}{3} + \binom{n_2}{3} + \dots + \binom{n_{2019}}{3}.$$

We first show $n_j \le 4$ for all $1 \le j \le 2019$. Indeed, if at least one is greater than $4$, then
$$2024 = \binom{n_1}{3} + \binom{n_2}{3} + \dots + \binom{n_{2019}}{3} \ge \underbrace{1+1+\dots+1}_{2018 \text{ times}} + \binom{5}{3} = 2028,$$
which is impossible.
Now suppose among all the $n_j$'s, $a$ of them are equal to $3$ and $b$ of them are equal to $4$. Then we have $a+b = 2019$ and
$$a + 4b = a\binom{3}{3} + b\binom{4}{3} = 2024.$$
This gives $b = \frac{(a+4b) - (a+b)}{3} = \frac{5}{3}$, which is again impossible. Therefore, such a choice does not exist.