This is how I would do it (although it has been 20 years since I last did electrics). Each load is made up of a resistance in series with an inductance. For the first load, it dissipates 50kW, which is resistive. You can work out the resistive element of each load by doing 50kW/120 (=416A), giving a resistance of 288m Ohms. Your power factor is 0.7, so you have the hypotenuse at 1 and the x axis at 0.7, so you need an inductance on the y axis that gives an angle of (cos^-1 of 0.7) (45.57 degrees) at 60 Hz (294 mOhms which is 779uH), so you have worked out the elements of the first load. Repeat this for the second load.
Work out the parallel combination of the 2 loads plus the capacitor. This 'lumped' load then forms a series network with the pure resistance that makes up the wire loss. You can then work out the current that flows through this and its phase with the input voltage.
I need to check the actual figs above, but that is the method I would use.
Others please discuss... and tell me where I have gone wrong...