A sound source is fixed at the origin and continuously emits sound...

The position of the sound source is at the origin $(0, 0)$.

The position of the listener is $(336, 140)$.

The distance between the source and the listener is:

$r = \sqrt{336^2 + 140^2} = \sqrt{112896 + 19600} = \sqrt{132496} = 364 \text{ m}$

Assuming the listener travels in the positive y-direction, the velocity of the listener is $\vec{v}_L = 26 \hat{j} \text{ m s}^{-1}$.

The unit vector along the line joining the source to the listener is:

$\hat{r} = \dfrac{336 \hat{i} + 140 \hat{j}}{364} = \dfrac{12}{13} \hat{i} + \dfrac{5}{13} \hat{j}$

The component of the listener's velocity along the line of sight (away from the source) is:

$v_o = \vec{v}_L \cdot \hat{r} = (26 \hat{j}) \cdot \left( \dfrac{12}{13} \hat{i} + \dfrac{5}{13} \hat{j} \right) = 26 \times \dfrac{5}{13} = 10 \text{ m s}^{-1}$

Since the listener is moving away from the source, the apparent frequency $f'$ is given by the Doppler effect formula:

f' = f_0 \left(\dfrac{v - v_o}{v} \right)

Substituting the given values:

f' = 660 \left(\dfrac{330 - 10}{330} \right) =660 \left(\dfrac{320}{330} \right) =$2 \times 320 = 640 \text{ Hz}$