Some Elementary Schubert Calculus Calculations 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 ?
 A: Edit. In the following parameter space, the "double line locus" $\Delta$, where $L$ equals $M$, appears multiply in the first intersection cycle $\delta \mu^2\nu^4\rho$.  Indeed, if $L$ equals $M$, if $p$ equals the intersection of $L=M$ with the specified $\rho$-hyperplane, and if $L=M$ is one of the $2$ lines that intersects each of the four specified $\nu$-lines, then the condition $\mu^2$ is "satisfied" for the corresponding points of $\Delta$.  This gives two "extra solutions" not desired by the OP, each with multiplicity $8$.  This accounts for the discrepancy between the number computed below, $33$, and the computation of the OP, $17$ (which also follows by the "duality involution" on the space of completed conics).  The five other cycles on the OP's list do not intersect $\Delta$.  
Edit. To address the OP's question in "Case 2": the set of lines in $3$-space that intersect each of two specified $\nu$-lines AND the line $N$ spanned by the two $\mu$-points is a smooth quadric surface $Q'$ containing $N$ -- what the OP calls a "regulus".  For the other two specified $\nu$-lines, there is a second smooth quadric surface $Q''$ containing $N$.  The common intersection of $Q'$ and $Q''$ is the union of $N$ and a twisted cubic curve $C$, a degree $3$, genus $0$ curve.  The intersection of $C$ with the $\rho$-hyperplane is $3$ points.  For each of these three points, there is a unique line in $Q'$, resp. in $Q''$, containing the point and intersecting $N$.  The union of those two lines is a conic.
Original answer. There are several descriptions of the parameter space of triples $(p,[L],[M])$ of a point $p$ in $3$-space, a line $L$ in $3$-space containing $p$, and a line $M$ in $3$-space containing $p$.  The (graded) Chow ring of these parameter space is, $$A^* = \mathbb{Z}[r,s,t]/I,\ \ I = \langle r^4,s^4,t^4,s^3+rs^2+r^2s+r^3,t^3+rt^2+r^2t+r^3 \rangle. $$ Here the degree $1$ class $r$ is the Chow class corresponding to the condition $\rho$.  The condition that $L$ intersect a specified line gives the degree $1$ Chow class $r+s$.  The condition that $M$ intersects a specified line gives the degree $1$ Chow class $r+t$.  The description of $A^*$ follows from Grothendieck's description of the Chow ring of a projective bundle.
The condition $\mu$ is $r+s+t$.  The condition $\nu$ is $(r+s)+(r+t)=2r+s+t$.  The socle of $A^*$, in degree $7$, is $r^3s^2t^2$.  Please note, this overcounts by a factor of $2$, since in each triple, the lines $L$ and $M$ are ordered.  For instance, this gives,
$$ \nu^6\rho = r(2r+s+t)^6 = $$ $$15\cdot 4\cdot r^3\cdot (6s^2t^2) + 6\cdot 2\cdot s^2\cdot (10t^3u^2+10t^2u^3) + s\cdot 20t^3u^3 = $$ $$15\cdot 4 \cdot 6 r^3 s^2 t^2 - 6\cdot 2 \cdot 10\cdot 2 r^3 s^2 t^2 + 20 r^3 s^2 t^2 = 140r^3s^2t^2,$$ instead of the true number, $70$.  Anyway, you can combine the presentation above for $A^*$ with any computer algebra program, such as Macaulay2, to compute the intersection numbers that you requested.  When I do this, I get the following answers for your enumerative problems,
$$ \begin{array}{cccc} \delta\mu^2\nu^4\rho & = & 33, & \textbf{N.B. } \text{Intersects } \Delta,\text{ see edit.}\\ \delta\mu\nu^4\rho^2 & = & 17, \\ \delta\mu\nu^6 & = & 70, \\ \delta\nu^6\rho & = & 70, \\ \delta \mu\nu^5\rho & = & 50. \end{array}$$
I posted the Macaulay2 script that I used to compute the intersection numbers at the following URL: http://www.math.stonybrook.edu/~jstarr/MO.m2
A: 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 $l''$.
