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 ?



Some Elementary Schubert Calculus Calculations
**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>


