The velocity of the plane relative to the air is $$250\hat{i}\text{ km/h}$$ . The velocity of the wind relative to the ground is $$75(\cos335^\circ\hat{i}+\sin335^\circ\hat{j})\text{ km/h}$$ $$=-52.5\hat{i}-43.3\hat{j}\text{ km/h}$$

Where $$\hat{i}\text{ and }\hat{j}$$ are the unit vectors in the x(east) and y(north) directions, respectively. The resultant velocity of the plane relative to the ground is $$\overrightarrow{v}_{pg}=\overrightarrow{v}_{pa}+\overrightarrow{v}_{ag}$$ $$=(250\hat{i}-52.5\hat{i}-43.3\hat{j})\text{ km/h}$$ $$=(197.5\hat{i}-43.3\hat{j})\text{ km/h}$$

The magnitude of the resultant velocity is

$$v_{pg}=\sqrt{(197.5)^2+(43.3)^2}$$ $$=\sqrt{39500+1875}$$ $$=\sqrt{41375}$$ $$\boxed{v_{pg}=203\text{ km/h}}$$

and the angle it makes with the x(east) axis is $$tan\theta\text{ tan}^{-1}\left(\frac{-43.3}{197.5}\right)$$ $$\theta=\boxed{-12.3^\circ }$$

So, the plane will fly at 203 km/h at 12.3 $^\circ$ south of east.