TITLE: Some Elementary Schubert Calculus Calculations
QUESTION [6 upvotes]: Here are some simple geometry problems I am unable to resolve to my satisfaction. I asked the question on Math Stack (https://math.stackexchange.com/questions/2713754/a-problem-in-elementary-combinatorial-space-geometry) but it has received no interest, so I ask here in a different format. 
Let $\delta$ denote the configuration of two intersecting lines, in other words, a conic with a double point, in three dimensional projective space. 
Let $\nu$ denote the condition that one of the lines in $\delta$ intersects some given line.
Let $\mu$ denote the condition that the plane of $\delta$ passes through a given point. 
Let $\rho$ denote the condition that the point of intersection of $\delta$ lies on a given plane. 
Prove:
$$\delta\mu^2\nu^4\rho=17$$
$$\delta\mu\nu^4\rho^2=17$$
$$\delta\mu\nu^6=70$$
$$\delta\nu^6\rho=70$$
$$\delta\mu\nu^5\rho=50$$
The first and second are the most baffling to me. The third and fourth, I have an idea, but I wish I had a better one.
Regarding the first, it means that the number of $\delta$ whose plane passes through two given points( $\mu^2$) and thus passes through a fixed line, and such that the lines of $\delta$ intersect $4$ given lines is $17$.
This can be analyzed as follows
${\bf Case 1}$ One of the lines of $\delta$ intersects three of the given lines. Then it intersects, these three lines and the axis of $\mu^2$, as ${\bf there \ are \ two \ lines \ intersecting \ four \ given \ lines \ in \ space}$ and there are 
$\binom{4}{3}=4$ such  choices there are $8$ lines. The second line is then uniquely determined.     
${\bf Case 2}$ This is the real problem. Each line of $\delta$ intersects $2$ of the given lines, as there are $\frac{1}{2}\binom{4}{2}=3$ such partitions 
each must have $3$ solutions, provided $17$ is the right answer. 
How to obtain this last calculation is the real problem. 
Well I have further thoughts, but I'll wait to see if any interest in this question. Am I missing something obvious ?

REPLY [4 votes]: I would like to propose a slightly more elementary derivation of $17$ in the language that you used on math.stackexchange. Indeed $17=2\cdot 4+3\cdot 3$.
So our hero: $l_1,l_2,l_3,l_4$ four lines, $l$ is one more line and $P$ is a plane. We are looking for coupes $l',l''$ so that $l'\cap l''\in P$, $l\subset P(l_1,l_2)$, and $l'\cup l''$ intersect all four $l_i$.
In your answer on math.stack you have explained why $2\cdot 4$ and why $3$. So only the last $3$ should be explained. Here is a claim:
There exists exactly $3=4-1$ couples $l',l''$ where $l'$ intersects $l_1,l_2$ and $l''$ intersects $l_3, l_4$.
Proof.  Note that the family of lines that intersect $3$ lines $l,l_1,l_2$ form a quadratic surface in $P^3$. Denote this surface $Q_{12}$. Denote by $Q_{34}$  the surface corresponding to lines $l, l_3,l_4$. Note now that $Q_{12}$ and $Q_{34}$ both intersect $P$ in conics. Denote these conics $C_{12}$ and $C_{34}$. By construction both $C_{12}$ and $C_{34}$ contain the intersection point $l\cap P$. And additionally to $l\cap P$ they have $3=4-1$ intersection points. Now, for each of these $3$ points we can pass two desired lines $l'$ and $l''$.