It sounds like you have a 50 HP motor, when all you need is a 10HP motor.

A new toy? Learn how to use it first. You might make sense of the readings then.

Wow! That's a horribly inefficient drive!

Measure the LRC and divide by 6,you will get approximate rated current.

whats that to mean?

If you apply P=sq rt 3 x V x I x cos phi,you will get power assuming reasonable value for cos phi & V or by measuring cos phi & V

Codswallop. The original poster knows nothing about the mechanical load that the motor is driving.

Use a dynamometer,tachometer,clampon ammeter,pf meter and voltmeter(power analyser) and keep on increasing the load and record the speed,V,I,pf and draw graphs of N,I,V,pf against Load, from which extrapolate the approximate power rating of the motor.

Check whether there are kW measured in both points - not kVar's. 7 kW lost in the drive is in any case too much. (I assume you do not have them in the form of scorching heat, deafening explosions, deadly radiation waves)

The rating of the motor is 37kW, because it is stated on the plate. It doesn't draw 37kW all the time. It will draw whatever the load requires, in terms of torque multiplied by angular velocity, up to a limit of 37kW. Above 37kW the motor is at risk, and the overload protection device(s) is/are there to protect it.

A measurement of 5kW going to a 37kW motor indicates that the motor is loaded to around 14% of its rating. This is no big deal, apart from the motor being 7 times bigger than the load needed at the time of the measurement, which may be either normal or an over-sizing error.

Since the motor is connected to a drive (VFD), shouldn't the drive be able to tell you the current power output to the motor making the use of the fluke redundant?