Cubic hypersurfaces through 5 generic lines in $\mathbb{P}^3$ Consider 5 generic lines $l_1, \dots, l_5 \subset \mathbb{P}^3$ (in particular, they do not intersect). Denote by $Z$ their union.
$$\dim H^0 \big( \mathbb{P}^3, \mathcal{O}_{\mathbb{P}^3} (3) \big) = \binom{3+3}{3} = 20$$ 
$$\dim H^0 \big( \mathbb{P}^3, \mathcal{O}_Z(3) ) = 5 \dim H^0( \mathcal{O}_{\mathbb{P}^1} (3) \big) = 5 \times 4 = 20$$
Consider restriction map $R$
$$R: H^0 \big( \mathbb{P}^3, \mathcal{O}_{\mathbb{P}^3} (3) \big) \rightarrow H^0 \big( \mathbb{P}^3, \mathcal{O}_Z(3) \big)$$
Question Is $R$ isomorphism?
Comment 1. You can reformulate this if you wish. Are there cubic hypersurfaces, passing through these 5 lines? What is $H^1 \big( I_Z(3) \big)$, where $I_Z$ - sheaf of ideals.
Comment 2. My question is particular case of this question. I tried to consider case by case. It is the easiest example, where I do not know the answer.
 A: Let’s prove that there is no cubic function, which vanishes on generic five lines $L_1, \dots, L_5$. Consider quadric $Q$ containing lines $L_1, L_2, L_3$ (it is a classical fact that for generic three lines there is unique quadric, containing them; moreover this quadric is smooth). 
Suppose there is a section $F \in H^0 \big( \mathcal{O}_{ \mathbb{P}^3 } (3) \big)$ which vanishes on $L_1, \dots, L_5$. Consider $f$ - restriction of $f$ to $Q$. Recall that $Q = \mathbb{P}^1 \times \mathbb{P}^1$. Let $x_1, x_2$ be homogeneous coordinates on first $\mathbb{P}^1$ and $y_1, y_2$ on the second one.
 $f(x_1, x_2 , y_1, y_2 )$ is a homogeneous polynomial of degree $(3, 3)$. Lines $L_1, L_2, L_3$ are sabvarietis of $Q$  defined as $a_i x_1 + b_i x_2 = 0$, $i= 1, 2, 3$. $f$ vanishes on these lines then
$$f(x_1, x_2, y_1 , y_2) = g( y_1, y_2) (a_1 x_1 + b_1 x_2)( a_2 x_1 + b_2 x_2)( a_3 x_1 + b_3 x_2)$$
So $f$ vanishes on three lines of type $\mathbb{P}^1 \times \{ (t_1: t_2) \}$ (denote them by $S_1$, $S_2$, $S_3$) and three lines of type $\{ (s_1: s_2) \} \times \mathbb{P}^1$ (namely $L_1$, $L_2$ $L_3$).
On the other hand $F$ vanishes on $L_ 4, L_ 5$. Then $f$ vanishes on their intersection with $Q$. It is 4 points (for generic position). These 4 points do not lie on $L_1, L_2, L_3$ (in generic position $L_i$ and $L_j$ do not intersect). So they lie on $S_1, S_2, S_3$. But 4 generic points of $Q$ do not lie on three lines. Then $f$ vanishes on $Q$.
So zero locus of $F$ is this quadratic $Q$ and a plane. Then $L_4$ and $L_5$ lies in this plane, which do not happens in generic position.
Appendix. Let's prove that for 3 non intersecting lines there is a unique quadric passing through them. And that it is smooth.
First of all let's see that there is one. Indeed $$\dim H^0 ( \mathcal{O}_{\mathbb{P}^3} (2) ) = \binom{2+3}{2} = 10$$
$$\dim H^0 ( \mathcal{O}_{Z} (2) ) = 3 \times 3 = 9$$
So we know that restriction map has at least one dimensional kernel. I leave it as an exercise to prove that any three lines on non smooth quadric intersects (prove this case by case for quadratic cone, union of two planes and double plane). So all quadrics passing through this lines are smooth. Another exercise is to prove that if kernel would have dimension strictly more than one, then there must be non smooth quadric. (hint: nonsmooth quadrics are characterized by equation $\det G =0$, where $G$ is matrix of symmetric form). This implies, that kernel is one dimensional. Equivalently, our quadric is unique.
