pairwise disjoint lines: $l_{1},l_{2},l_{3}, l_{4} \in \mathbb{P}(\mathbb{R}^{4})$

Suppose I have 4 pairwise disjoint lines: $l_{1},l_{2},l_{3}, l_{4} \in \mathbb{P}(\mathbb{R}^{4})$. I want to find the number of lines, which intersect all these 4 lines. Is the number of lines stable to "small" movments of $l_{1},l_{2}, l_{3}, l_{4}$.

$\exists !$ quadric $Q \subset \mathbb{P}^{3}:l_{1}, l_{2}, l_{3} \subset Q $ and planarity $Q$ is equal to 1. So, $Q$ is congruous to Segre quadric and the answer is $0,1,2$ or infinity. But I can't understand is it stable to "small" movements... give me any ideas please

Thank you in advance!

Solution 1:

Once you have the quadric $Q$ containing $l_1,l_2,l_3$ (which are pairwise skew), then the question is: In how many points does $l_4$ intersect $Q$? If that number is $0$ or $2$, a small perturbation will not change the intersection number of the line with $Q$. The number cannot in fact be $1$, because any line that is tangent to the quadric must be wholly contained in it. But a general small perturbation of a line contained in the quadric will no longer lie in the quadric.