Let $L,M,N \in \mathbb{P}^5$ $3$ general subspaces of codimension $3$, and let $l_i$ (resp $m_i,n_i)$, $i=1,2,3$ be three general points on $L_i$ (resp. $M_i,N_i)$.
I can't understand why $I_{L\cup M\cup N}(3)$, (i.e. the space of all cubics through these subspaces with all first partial derivatives vanishing at this $9$ points), has dimension $26$.
I really can't figure out how to attack the problem, so any hint would be appreciate!
The dimension of the space of cubics passing through $L$, $M$, and $N$ without extra first derivative conditions at $l_i$, $m_i$, and $n_i$ is 26. And if you impose these extra conditions, the dimension will drop to 17.
To see this note that for a general triple $L,M,N$ there is a unique $\mathbb{P}^2 \times \mathbb{P}^1 \subset \mathbb{P}^5$ such that $$ L = \mathbb{P}^2 \times \{0\}, \quad M = \mathbb{P}^2 \times \{1\}, \quad N = \mathbb{P}^2 \times \{\infty\}. $$ Cubics in $\mathbb{P}^5$ cut out a complete linear system $\mathcal{O}(3,3)$ and the space of cubics containing $\mathbb{P}^2 \times \mathbb{P}^1$ has dimension 16.
The ideal of $L \cup M \cup N$ in $\mathbb{P}^2 \times \mathbb{P}^1$ is $\mathcal{O}(0,-3)$, so the space of sections of $\mathcal{O}(3,3)$ vanishing on $L \cup M \cup N$ coincides with $H^0(\mathbb{P}^2 \times \mathbb{P}^1,\mathcal{O}(3,0))$, hence has dimension 10. Altogether, this gives $$ 16 + 10 = 26 $$ cubics through $L$, $M$, $N$.
If you want to impose the first derivative conditions at nine points $l_i$, $m_i$, $n_i$, this means that the above section of $\mathcal{O}(3,0)$ vanishes at the corresponding nine points of $\mathbb{P}^2$, and such section is unique. This gives $$ 16 + 1 = 17 $$ as the answer to this question.