Homotopy equivalence of $\mathbb{R}^3 \setminus \{$lines$\}$ I'm aware that $\mathbb{R}^3 \setminus \{$a single line$\}$ is homotopy equivalent to $\mathbb{R}^2 \setminus \{$pt$\}$. Similarly, $\mathbb{R}^3 \setminus \{$two parallel lines$\}$ is homotopy equivalent to $\mathbb{R}^2 \setminus \{$two points$\}$. However, I'm not sure how to find homotopy equivalences for the following cases:


*

*$\mathbb{R}^3 \setminus \{$two disjoint, non-parallel lines$\}$

*$\mathbb{R}^3 \setminus \{$two lines intersecting at a point$\}$

*The above cases but with $n$ lines instead of $2$.
If anyone is able to describe the necessary deformation retracts or point me in the direction of something that does, that would be very much appreciated. My end goal is to calculate the de Rham cohomology of the above spaces, but I'm only aware of homotopy equivalences/Mayer-Vietoris sequence for doing such things.
 A: Let me address items 1 and 3 in a cetain amount of detail, and item 2 very briefly at the end. 
You say you are aware of homotopy equivalences and Mayer-Vietoris, but don't neglect homeomorphisms!
First, from what you are aware regarding $\mathbb R^3 \setminus \{\text{a single line}\}$ I'm sure you could figure out that $\mathbb R^3 \setminus \{\text{$n$ parallel lines all in the same plane}\}$ is homotopy equivalent to $\mathbb R^2 \setminus \{\text{$n$ points all in the same line}\}$.
The key idea is to prove:

The homeomorphism type of $\mathbb R^3 \setminus \{\text{$n$ pairwise disjoint lines}\}$ is well-defined, depending only on $n$. 

So in your case 1, the space is homeomorphic to $\mathbb R^3 \setminus \{\text{two disjoint, parallel lines}\}$ which is homotopy equivalent to $\mathbb R^2 \setminus \{\text{two points}\}$.
Let me do the case $n=2$ in full detail, after which I'll suggest some ideas for going beyond that case.

Lemma: $R^3 \setminus \{\text{two disjoint non-parallel lines}\}$ is homeomorphic to $\mathbb R^3 \setminus \{\text{two disjoint parallel lines}\}$.

Consider two disjoint non-parallel lines $L,M$. Connect them by the unique line segment $\overline{A B}$, $A \in L$, $B \in M$ that means $L$ and $M$ at right angles. Let $P$ be the plane that bisects $A,B$ perpendicularly. By applying a rigid motion of Euclidean space, we may assume that $P$ is the $xy$-plane, $A = (0,0,c)$, and $B = (0,0,-c)$. The line $L$ lies in the plane $z=c$, and by applying a rigid rotation of Euclidean space around the $z$-axis we may assume that $L$ is parallel to the $x$-axis. Looking down the $z$-axis and projection to the $x,y$ plane, the projection of $L$ is the $x$-axis, and the projection of $M$ makes some positive angle $\theta$ with respect to the $x$-axis.
Now we are going to apply a non rigid homeomorphism of Euclidean 3-space, in three parts.
First, all points with $z \ge 0$ are fixed. 
Second, the entire portion of the plane with $z \le -c$, is rigidly rotated around the $z$ axis by an angle of $-\theta$; after this rotation, the image $M'$ of the line $M$ is parallel to the $x$-axis hence to $L$. 
Third, for $-c \le z_0 \le 0$, the plane $z=z_0$ is rotated rigidly around the $z$ axis by an angle $-\theta \cdot \frac{z_0}{c}$.
That's it for $n=2$!
I think the best thing to do in general is to use the ideas of this proof to construct a proof by induction for larger values of $n$.
For some final comments regarding item 2, instead of planes, think of spheres centered at the intersection point $P$: the space $\mathbb R^3 - \{\text{two lines intersecting at $P$}\}$ is homotopy equivalent to a sphere centered on $P$ minus the four points where those lines puncture that sphere.
