How to show that $\mathbb{R}^3$ with two disjoint parallel lines removed has the same homotopy type as $N$.

Question

I am studying for an exam and came across a problem from a previous incarnation.

The problem is stated as follows:

Let $L_1$ and $L_2$ be disjoint parallel lines in $\mathbb{R}^3$. Let$ M=\mathbb{R}^3−(L_1\cup L_2)$. Let $q\in M$. Let $C_1$ $C_2$ be disjoint parallel lines in $\mathbb{R}^3$. Let $N=\mathbb{R}^3−(C_1\cup C_2)$.

1. Show that $\pi_1 M\cong \pi_1N$.
2. Show that $M$ and $N$ do not have the same homotopy type.
3. Show that $M−\{q\}$ has the same homotopy type as $N$.

First, I am farily certain that $C_1$ and $C_2$ should be circles rather than lines. I am assuming that $C_1$ and $C_2$ are disjoint and not linked.

1. I was able to show that $M$ and $N$ have the same fundamental group by first showing that $M$ deformation retracts to a bouquet of two circles. Similarly, $N$ is homotopy equivalent to $S^2\vee S^2 \vee S^1\vee S^1$. I can then use the Seifert van Kampen theorem to compute the fundamental groups.

2. For the second part, I computed the Euler characteristic of $M$ and $N$. Seeing as $\chi(M) = -1$ while $\chi(N) = 1$, these spaces are not homotopy equivalent.

3. This is where I get stuck. I already know that $M$ is homotopy equivalent to the bouquet of two circles, so I was hoping to show the same for $N$. However, I am unable to do this. Any ideas on how to approach this problem?