I can't seems to get the right answer and I totally don't know how to do it. Please someone help me.

In Figure Q1.2, a three-phase four-wire system operating with a line voltage of 480 V is supplying a balanced motor load at 560 kVA and 0.86 power factor (pf) lagging. The motor load is connected to the three main lines marked a, b, and c. In addition, incandescent lamps (with unity pf) are connected as follows: 48 kW from line a to the neutral, 25 kW from line b to the neutral, and 14 kW from line c to the neutral. The three-wattmeter method is used to measure the power absorbed by the load in each line.
(I) Calculate the wattmeter readings.
(ii) Find the current in the neutral line.

This is the figure:
http://i486.photobucket.com/albums/rr230/Azwaney/figure.jpg

First you convert the motor kVA to kW.

Can you do that?

Since the load presentred by the motor is given as 560 kVA at 0.86 power factor, to get the kW drawn by the motor, it is calculated as follows: 560 kVA * 0.86 = 481.6 kW. Divde this by 3 to get the kW supplied by each phase which equals 160.5 kW. To get the kW loading for each phase then add the kW loads on each phase to get, a = 208.5 kW, b=185.5 kW, c=174.5 kW.

Last the neutral current is found by adding the vector currents from each phase to get the resultant. At first this looks a bit complicated due to the power factor of the motor load, however, the motor load is balanced and thus it will have zero neutral current addition. This leaves only the resistive loads at unity power factor, thus the phase angle between each current will be 120 degrees if we assume zero impedance from the source and zero impedance for the wires supplying the load. Then we can find each current by dividing the single phase loads by 277 volts to get, a = 173 amps at 0 degrees, b=90 amps at 240 degrees and c = 51 at 120 degrees. Perform the vector sum of these currents to get a=173+j0 amps, b=-45-j78 amps, c=-25+j44 amps. Add these to get the neutral current of 103-j34 amps. Convert to polar notation to get 108 amps @-18 degrees. Hope this helps,
Newton1Law