Given that \( A B C, D E F \) and \( O G H \) are equally spaced parallel lines, and \( A D O, B E G \) and \( C F H \) are also equally spaced, we can use the information provided to express the vectors in terms of \( \mathbf{a} \) and \( \mathbf{h} \).
1. \(\overrightarrow{O B} = \mathbf{a} \): Since \( O B \) lies on line \( O G H \), which is parallel to \( A B C \), it is aligned with the vector \( \mathbf{a} \).
2. \(\overrightarrow{H D} = -\frac{1}{2} \mathbf{a} + \mathbf{h} \): The vector \( \overrightarrow{H D} \) can be expressed as a combination of \( \mathbf{a} \) and \( \mathbf{h} \). Since \( P \) is the midpoint of \( A D \), \( \overrightarrow{A P} = \frac{1}{2} \overrightarrow{A D} \), and \( \overrightarrow{H D} = -\overrightarrow{A P} + \overrightarrow{A H} = -\frac{1}{2} \mathbf{a} + \mathbf{h} \).
3. \(\overrightarrow{F A} = \frac{3}{2} \mathbf{a} - \mathbf{h} \): Similar to the reasoning above, \( \overrightarrow{F A} = \overrightarrow{F O} + \overrightarrow{O A} = \frac{3}{2} \mathbf{a} - \mathbf{h} \).
4. \(\overrightarrow{A F} = -\frac{3}{2} \mathbf{a} + \mathbf{h} \): Since \( \overrightarrow{A F} = -\overrightarrow{F A} \), we can write \( \overrightarrow{A F} = -\frac{3}{2} \mathbf{a} + \mathbf{h} \).
