I am having a problem with the 200 degrees flight also, but, ....

Imagine, or draw the figure on paper.

Flight directions are measured from positive y-axis going clockwise.

If we call the quadrants in the usual way,

320 degrees is in the 2nd quadrant,

200 degrees is in the 3rd quadrant.

The angle from 200deg to 320deg is

320 -200 = 120 degrees.

distance = rate * time

So, after 3 hrs,

150 km/h *3h = 450 km

200 km/h *3h = 600 km

Hence, the figure is an obtuse triangle ABC where

side AB = 450 km

angle ABC = 120 degrees

side BC = 600 km

side CA = distance between the two planes

That is two known sides and their included angle.

By Law of Cosines,

(CA)^2 = (AB)^2 +(BC)^2 -2(AB)(BC)cos(120deg)

(CA)^2 = (450)^2 +(600)^2 -2(450)(600)cos(120deg)

(CA)^2 = 202,500 +360,000 -540,000(-0.5)

(CA)^2 = 202,500 +360,000 +270,000

(CA)^2 = 832,500

CA = sqrt(832,500)

CA = 912.4

Therefore, after 3 hours, the two planes are 912.4 km apart. ---answer.