Is the following result about congruences of lines classical? The following result must be classical; I am looking for a reference or an easy proof (I have a rather sophisticated argument).
Let $\mathbb{G}$ be the Grassmannian of lines in $\mathbb{P}^3$, and let $S$ be the complete intersection of two general hypersurfaces in $\mathbb{G}$, of fixed degrees $>1$. Then $S$ parametrizes a 2-dimensional family of lines (a congruence in classical language). The claim is that through any point of $\mathbb{P}^3$ pass only finitely many such lines (in classical language again, the fundamental locus is empty). 
 A: Let $V = \mathbb{C}^4$ and let $a \in V$. The set of $\mathbb{C}^2 \subset V$ which passes through $a$ is a $\mathbb{P}^2$ linearly embedded in $G(2,V) \subset \mathbb{P}(\bigwedge^2 V)$. The Grassmannian $G(2,4) \subset \mathbb{P}(\bigwedge^2 V)$ is identified with a quadric and the $\mathbb{P}^2$ described above is a maximal isotropic subspace of $\mathbb{P}(\bigwedge^2 V)$ for the corresponding quadratic form. When $a$ varies in $\mathbb{P}^3$, we get a $3$-dimensional family of isotropic subspaces of $ \mathbb{P}(\bigwedge^2 V)$. This family is identified with the spinor variety of isotropic $\mathbb{C}^3 \subset \bigwedge^2 V$. The spinor variety is in this case isomorphic to $\mathbb{P}^3$, but for reading convenience, we denote it by $\mathrm{Spin}^{3}$.
Let $\mathcal{R}$ be the taulogical bundle of  $\mathrm{Spin}^{3}$. We consider $ p :X = \mathbb{P}(\mathcal{R}) \longrightarrow \mathrm{Spin}^3$. By the Borel-Bott-Weyl Theorem, we have an identification (to check?) :
$$ H^0(X, \mathcal{O}_{X/\mathrm{Spin}^3}(d)) = H^0(\mathrm{Spin^3}, S^d \mathcal{R}^*) = S^d (\bigwedge^2 V)^* =  H^0(\mathbb{P}(\bigwedge^2 V), \mathcal{O}(d)).$$
We deduce that a pair of general hypersurfaces of degree $d$ and $e$ in $\mathbb{P}(\bigwedge^2 V)$ can be seen as a generic section of $S^d \mathcal{R}^* \oplus S^e \mathcal{R}^*$ on $\mathrm{Spin}^3$ and can be also seen as a generic section of $\mathcal{O}_{X/\mathrm{Spin}^3}(d) \oplus \mathcal{O}_{X/\mathrm{Spin}^3}(e)$. 
Your statement can now be reformulated as follows :

Statement: Let $d,e \geq 2$ and $f \in H^0(X,\mathcal{O}_{X/\mathrm{Spin}^3}(d)  \oplus \mathcal{O}_{X/\mathrm{Spin}^3}(e))$ be a general section. For all $u \in \mathrm{Spin}^{3}$, the scheme $\{f = 0\} \cap p^{-1}(u)$ is of codimension $2$ in $p^{-1}(u)$.

This is looks very like a Bertini-type result. In that direction, one can prove the following :

Lemma 1 : Let $e \geq 2$ and general $h \in H^0(X,\mathcal{O}_{X/\mathrm{Spinor}_3}(e))$. Then, for all $u \in \mathrm{Spin}^{3}$, the scheme $\{h = 0\} \cap p^{-1}(u)$ is of pure codimension $1$ in $p^{-1}(u)$.
proof: the section $h$ induces a section a generic section of $S^e \mathcal{R}^*$. The vanishing locus of this section corresponds to the variety of isotropic $\mathbb{P}^2 \subset \mathbb{P}(\bigwedge^2 V)$ which are included in the hypersurface in $\mathbb{P}(\bigwedge^2 V)$ corresponding to $h$. 
   We only have to check that the vanishing locus of this section is empty. We have $\dim S^e \mathcal{R}^* = \binom{e-1+3}{e}> \dim \mathrm{Spin}^{3}$ when $e \geq 2$. Furthermore, the vector bundle $\mathcal{R}^*$ is globally generated, we deduce that the vanishing locus of $h$ is empty.
Lemma 2 [see here, the idea is due to Zak] Let $X \subset \mathbb{P}^N$ be an equidimensional scheme of dimension $n$. Denote by $\mathcal{H}_{N,e}$ the projective space of hypersurfaces of degree $e$ in $\mathbb{P}^N$. Then, the set of points in $\mathcal{H}_e$ which contains an irreducible component of $X$ is of codimension at least $\binom{e+n}{e}$.
proof : [see here]. 

Using Lemma $1$ and $2$, one proves the following:

proposition : Let $e \geq 2$, $d \geq 3$ and $(g,h) \in H^0(X,\mathcal{O}_{X/\mathrm{Spin}^3}(d)  \oplus \mathcal{O}_{X/\mathrm{Spin}^3}(e))$ be a general section. For all $u \in \mathrm{Spin}^{3}$, the scheme $\{(g,h) = 0\} \cap p^{-1}(u)$ is of codimension $2$ in $p^{-1}(u)$.
proof : Denote by $X_h$ the vanishing locus of $h$ in $X$. By Lemma $1$, we know that all the fibers of the map:
  $$ p_h : X_h \longrightarrow \mathrm{Spin}^3$$
  are of pure dimension $1$. Let $\mathcal{H}_{\bigwedge^2V,d}$ be the projective spaces of hypersurfaces of degree $d$ in $\mathbb{P}(\bigwedge^2 V)$. Denote by $J \subset \mathcal{H}_{\bigwedge^2V,d} \times S$ the subvariety defined by:
  $$ J = \{(H,s) \in \mathcal{H}_{\bigwedge^2V,d} \times S, H \ \textrm{contains an irreducible component of} \ p_h^{-1}(s) \}$$
  By Lemma $2$, we know that each fiber of $q : J \longrightarrow S$ has codimension at least $\binom{d+1}{d}$ in $\mathcal{H}_{\bigwedge^2V,d}$. As a consequence, the codimension of $J$ in $S \times \mathcal{H}_{\bigwedge^2V,d}$ is at least $\binom{d+1}{d}$.
  Since $\dim S = 3$, we get that $\mathrm{codim}(J) - \dim S - 1 \geq 0$ for $d \geq 3$. Hence, the map $p : J \longrightarrow \mathcal{H}_{\bigwedge^2V,d}$ can't be dominant.
  We deduce that for generic $g \in \mathcal{H}_{\bigwedge^2V,d}$, the scheme $\{(g,h) = 0\} \cap p^{-1}(s)$ has dimension $0$ in $p^{-1}(s)$, for all $s \in S$.

