Given:
- Three phase, three wire system
- Two wattmeters reading 4000 Watts and 2000 Watts respectively

To find:
- Power factor when both meters give direct reading

Solution:
In a three phase, three wire system, the power is given by the formula:
P = √3 * V * I * cos(θ)

Where:
- P is the power in Watts
- V is the line-to-line voltage in volts
- I is the line current in amps
- θ is the phase angle between the voltage and current

When both wattmeters give direct reading, it means that one wattmeter is connected in one of the line wires and the other wattmeter is connected in one of the other line wires. The third line wire is not connected to any wattmeter.

Let's assume that the wattmeter reading 4000 Watts is connected in line 1 and the wattmeter reading 2000 Watts is connected in line 2.

From the given information, we know that:
Wattmeter 1 reading = 4000 Watts
Wattmeter 2 reading = 2000 Watts

The total power in the system can be calculated by adding the readings of both wattmeters:
Total power = 4000 Watts + 2000 Watts = 6000 Watts

Since the third line wire is not connected to any wattmeter, it means that the power flowing through it is zero.

Applying the power formula for each wattmeter individually, we can find the line currents (I1 and I2) and the phase angles (θ1 and θ2) for each wattmeter.

For Wattmeter 1:
4000 = √3 * V * I1 * cos(θ1)

For Wattmeter 2:
2000 = √3 * V * I2 * cos(θ2)

As the power factor is the same for both wattmeters, we can assume that the phase angles (θ1 and θ2) are equal.

Now, let's find the power factor using the given values.

1. Calculate the total power:
Total power = 6000 Watts

2. Calculate the power factor:
Power factor = Total power / (√3 * V * I1 * cos(θ1))
= 6000 / (√3 * V * I1 * cos(θ1))

Since the power factor is a ratio of real power to apparent power, it is always between 0 and 1. Therefore, the power factor cannot be 1 or greater than 1.

Hence, the correct answer is option C) 0.866.