Infinitely many equidistant parallel lines are drawn in the plane. A positive integer $n \geqslant 3$ is called frameable if it is possible to draw a regular polygon with $n$ sides all whose vertices lie on these lines, and no line contains more than one vertex of the polygon.

(a) Show that $3, 4, 6$ are frameable.
(b) Show that any integer $n \geqslant 7$ is not frameable.
(c) Determine whether $5$ is frameable.

[i]Proposed by Muralidharan[/i]
1. **Setup and Definitions:**
   - Suppose the plane consists of infinitely many equidistant parallel lines.
   - Assume these lines are parallel to the real axis and are spaced one unit apart.
   - Let \( P_0, P_1, \ldots, P_{n-1} \) be the vertices of an \( n \)-sided regular polygon. Assume \( P_i = z \cdot \zeta^i \) for \( i \in \{0, 1, 2, \ldots, n-1\} \), where \( z \) is a complex number and \( \zeta = e^{2\pi i / n} \) is the \( n \)-th root of unity.
   - Denote the parallel lines as \( l_i \) such that \( P_i \in l_i \).

2. **Imposing Conditions:**
   - The imaginary part of \( P_i \) must lie on the \( i \)-th line. Thus, \( \text{Im}(P_i) = y \cos \left( \frac{2\pi i}{n} \right) + x \sin \left( \frac{2\pi i}{n} \right) \).
   - For the lines to be equidistant and parallel, the imaginary parts must differ by integers.

3. **Rationality Condition:**
   - From the above, we get:
     \[
     l_0 \equiv y \quad \text{(1)}
     \]
     \[
     l_1 \equiv y \cos \left( \frac{2\pi}{n} \right) + x \sin \left( \frac{2\pi}{n} \right) \quad \text{(2)}
     \]
     \[
     l_{n-1} \equiv y \cos \left( \frac{2\pi (n-1)}{n} \right) - x \sin \left( \frac{2\pi (n-1)}{n} \right) \quad \text{(3)}
     \]
   - From (2) and (3), we get that \( x \sin \left( \frac{2\pi}{n} \right) \) and \( y \cos \left( \frac{2\pi}{n} \right) \) must be rational numbers.
   - From (1) and (2), we get that \( y \) must also be a rational number, implying \( \cos \left( \frac{2\pi}{n} \right) \) is rational.

4. **Part (a): Show that \( 3, 4, 6 \) are frameable:**
   - For \( n = 3 \):
     \[
     \cos \left( \frac{2\pi}{3} \right) = -\frac{1}{2} \in \mathbb{Q}
     \]
   - For \( n = 4 \):
     \[
     \cos \left( \frac{2\pi}{4} \right) = 0 \in \mathbb{Q}
     \]
   - For \( n = 6 \):
     \[
     \cos \left( \frac{2\pi}{6} \right) = \frac{1}{2} \in \mathbb{Q}
     \]
   - Therefore, \( n = 3, 4, 6 \) are frameable.

5. **Part (b): Show that any integer \( n \geq 7 \) is not frameable:**
   - We use the fact that \( 2 \cos \left( \frac{2\pi}{n} \right) \) is a root of the Chebyshev polynomial \( T_n(x) \) of degree \( n \).
   - If \( 2 \cos \left( \frac{2\pi}{n} \right) \) is rational, then \( T_n(x) - 2 \) has a rational root.
   - The only rational values for \( \cos \left( \frac{2\pi}{n} \right) \) that satisfy this are \( \pm 1 \) and \( \pm \frac{1}{2} \), corresponding to \( n = 1, 2, 3, 4, 6 \).
   - For \( n \geq 7 \), \( \cos \left( \frac{2\pi}{n} \right) \) is not rational, hence \( n \geq 7 \) are not frameable.

6. **Part (c): Determine whether \( 5 \) is frameable:**
   - For \( n = 5 \):
     \[
     \cos \left( \frac{2\pi}{5} \right) = \frac{\sqrt{5} - 1}{4} \notin \mathbb{Q}
     \]
   - Therefore, \( n = 5 \) is not frameable.

The final answer is \( \boxed{ n = 3, 4, 6 } \) are frameable, \( n \geq 7 \) are not frameable, and \( n = 5 \) is not frameable.