Assume that such configuration can be obtained. Let $n_i$ be the number of the points in the $i^{th}$ plane. Let $F$ be the set of all quadruples $(\{i,j,k\},l)$ s.t $1 \leq i<j<k \leq 24 , \ 1 \leq l \leq 2013$ and the $i^{th},j^{th},k^{th}$ points lie on the $l^{th}$ plane. Now we double count the size of this set. Every three points determine exactly one plane (as every three points are non-collinear), so $|F| = \binom{24}{3} = 2024$. Now every plane contain $n_i$ points, hence $\binom{n_i}{3}$ triples of points, hence $|F| = \sum_{i=1}^{2013} \binom{n_i}{3}$. So $$ 2024 = \sum_{i=1}^{2013} \binom{n_i}{3} $$. We note that $n_i \leq 5$, as otherwise, the sum on the right side is $\geq \binom{6}{3} + 2012 = 2032$ , which is impossible. Now note that $3 | \binom{n_i}{3} - 1$, for $n_i \in \{3,4,5\}$, so $3$ divides $$ \sum_{i=1}^{2013} \left(\binom{n_i}{3} - 1 \right) = 2024-2013 =11 $$which is absurd.