Intersection of lines in $ \mathbb{P}^3 $ can be given as zero locus of homogeneous linear polynomial in $ \mathbb{P}^5 $ I'm currently stuck at Exercise 8.19 b) in the notes to Algebraic Geometry from  Gathmann

Let $L \subset \mathbb{P}^3$ be an arbitrary line. Show that the set of lines in $\mathbb{P}^3$ that intersect $L$, considered as a subset of $G(2,4) \subset \mathbb{P}^5 $, is the zero locus of a homogeneous linear polynomial.

I tried solving the problem by first fixing a line in $\mathbb{P}^3$ and looking at all lines that intersect it and then derive the form of the needed linear homogeneous linear polynomial. But have no idea how.
So i started with fixing $L$ as $Lin(e_1 ,e_2) $ this leads to coordinate $(1:0:0:0:0:0)$ in $\mathbb{P}^5$ with Plücker embedding. Obviously the lines $(e_1,e_3),(e_1,e_4),(e_2,e_3),(e_2,e_4),(e_3,e_4)$ intersect $L$ and have coordinates $(0:1:0:0:0:0),...,(0:0:0:0:1:0)$.
Now i need to also include lines that are linear combinations of my basis vectors for example $(e_1+e_2,e_2+e_3)$. My question is how to find all of these lines and how to build a homogeneous linear polynomial from them.
 A: Let $L$ be the line given by the projectivization of the span of $e_1, e_2$. In other words, $L$ is the vanishing locus of $e_3$ and $e_4$. Let $v, w \in (\mathbb{C}^4)^*$ be linearly independent dual vectors (i.e., homogeneous linear forms, whose vanishing loci are planes), and let $L'$ be the line whose image in the Plucker embedding is given by $v \wedge w$ (i.e., $L'$ is the vanishing locus of the functionals $v,w$). Then $L \cap L'$ is nonempty precisely when there exists $[a : b ] \in \mathbb{P}^1$ such that $v(ae_1 + be_2) = w(ae_1 +be_2) = 0$. Let $v = \sum_{i = 1}^4 c_ie_i^*$ and $w = \sum_{i = 1}^4 d_ie_i^*$. Then $v(ae_1 + be_2) = w(ae_1 +be_2) = 0$ is equivalent to saying that $ac_1 + bc_2 = ad_1 + bd_2 = 0$, so $L \cap L'$ is nonempty precisely when the matrix $\left[\begin{array}{cc} c_1 & c_2 \\ d_1 & d_2\end{array}\right]$ has vanishing determinant, which occurs when $c_1d_2 - d_1c_2 = 0$. But the $e_1^* \wedge e_2^*$ coordinate of $v \wedge w$ is given by $c_1d_2 - d_1c_2$, so $L \cap L'$ is nonempty precisely when $L'$ has vanishing $e_1^* \wedge e_2^*$ coordinate under the Plucker embedding, and this is obviously a hyperplane condition on the $\mathbb{P}^5$ in which $G(2,4)$ sits.
