Assume $k$ is algebraically closed field of char zero. Let $l_1,l_2,l_3$ be three disjoint lines in $\mathbb{P}^3_k $. The dimension of the $k$-vector $\operatorname{Homog}_2 k[x_0, x_1, x_2, x_3]$ of degree $2$ homogeneous polynomials in $4$ variables is $10$ by usual Stars and Bars trick. We can associate to every such homogeneuos polynomial $P_2(x)$ of degree $2$ a hypersurface $H_{P_2} \subset \mathbb{P}^3_k $ which equals on each affine chart $D(L)$ with linear polynomial $L= \sum a_i x_i$ the vanishing set of points of the obtained polynomial after localization in $L$.
Question: Is there a combinatorical argument (in spirit of stars and bars trick) why the subvector space of $\operatorname{Homog}_2 k[x_0, x_1, x_2, x_3]$ consisting of all homog polynomials of degree $2$ which have associated hyperplanes containing all three disjoint lines $l_1,l_2,l_3$ is non zero.
Can it be derived more cenceptionally?
Say we have an arbitrary point $p \in \mathbb{P}^3_k $ and a line $l \subset \mathbb{P}^3_k $. What is the dimension of subspace of $\operatorname{Homog}_2 k[x_0, x_1, x_2, x_3]$ which have associated hyperplanes containing point $p$? Or the line $l$? Or both? If we know it how to approach combinatorically this problem we can also ask the natural question about the dimension of subspace of homog polynomials which have hyperplanes contain two oder more points, lines or more dimensional linear subspaces of $\mathbb{P}^3$?
Similary can such combinatorial argument be generalized to say $\mathbb{P}^n $, $\operatorname{Homog}_m k[x_0, x_1, ..., x_n]$ with $m \le n$ and one asks what is the dimension of $k$-subspace of polynomials which have associated hyperplanes all containing certain linear subspace $L_1 \subset \mathbb{P}^n $ or some subspaces $L_1,..., L_r \subset \mathbb{P}^n $ simultaneously?
This is a beautiful subject that gets hard very quickly. To answer your initial question: notice that if we fix a line $\ell \subset \mathbb P^3$, it meets every hyperplane; the intersection is either a point, or the entire line (obviously this latter case is when the line lies in the hyperplane). On the other hand, if we fix a point $p \in \mathbb P^3$, there is a codimension $1$ subspace of hyperplanes which contain $p$ (in the Grassmannian/dual projective space $\mathbb P^{3\vee} = \mathbb G(2,3) = G(2,4)$ of hyperplanes in $\mathbb P^3$). Since a different point $q$ determines a different codimension $1$ subspace, their intersection has codimension $2$. A third point results in a codimension $3$ linear space, which in $\mathbb P^3$ is a point: in other words, $3$ general points determine a $2$-plane, which you've known since high school.
The space of quadrics containing a given line $L$ can be computed by supposing that we have homogeneous coordinates $(W:X:Y:Z)$ on $\mathbb P^3$ such that $L$ is $W=X=0$. Then the equation for such a quadric has the form $Q = W\cdot L_1(W,X,Y,Z) + X \cdot L_2(W,X,Y,Z)$ where $L_1,L_2$ are linear forms. The (projective) space of linear forms is $3$-dimensional, so overall you have a $6$-dimensional space of quadrics containing $L$. This has codimension $3$ in the $9$-dimensional projective space of quadrics. Choosing two other general lines, you get two more codimension $3$ subspaces, so the intersection is codimension $9$ (codim. is additive over intersections of generic subspaces) a.k.a. dimension $0$, hence we get not just non-emptiness of the space of quadrics containing the three lines, but in fact that the quadric is unique.