Complex power calculation across parallel loads
You start by finding the impedance for each of the loads.

Load 1: 

$$ I_{1}=\frac{20\times 10^{3}}{600}\angle -36.8 = 33.33A \angle -36.8 $$
$$ Z_{1}=\frac{V}{I_{1}}= \frac{600}{33.33\angle-36.8}=18\angle36.8$$

Load 2:

$$Z_{2}=R_{2}\left |  \right |X_{2} $$

$$Z_{2}=((15)^{-1}+(j30)^{-1})^{-1}=13.42\angle 26.565$$

Load 3:

$$\cos(\tan^{-1}(\frac{X_{3}}{R_{3}}))=0.6$$
$$X_{3}=3.33R_{3}$$

Now consider all 3 loads in parallel and reduce it to a single impedance, you should get this in terms of one unknown variable.Then you know the overall power factor of the reduced load so solve it for one unknown variable R3 or X3.

From that point onwards you know the voltage drop across Z3 and hence I3, everything else derives itself as @Andyaka stated.