**This answer was edited. If needed, see history**

It is known that the sum of the two wattmeters is the total power of the load, but it doesn't mean that both of them will present the same value, since the **sources** are **out of phase**. As noted by Transistor, the wattmeters read the lines in sequence.

Found a document with an exercise. Refer to example 20.1 or 20.3

https://nptel.ac.in/content/storage2/courses/108105053/pdf/L-20(NKD)(ET)%20((EE)NPTEL).pdf

Also, from the same document:

```
PF=1 → W1=W2
PF=0.5 (cos60) → W1=Total power
0.5<PF<1 → W1>W2
```
----------------------------------------------------------
**For the proposed problem:**

Total power (all three loads):

```
P=|VL|²/R*cos(<Z)
OR
P=3*|Vp|*|Ip|*cos(<Z)
```

Line power:

```
W1=|VL|*|IL|*cos(30+(<Z))
W2=|VL|*|IL|*cos(30-(<Z))
```

In my case:

```
W1= 380*220/10*cos(30+60)= 0
W2= 380*220/10*cos(30-60)= 7240
```

This answer is free to be improved