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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
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*|Vf|*|If|*cos(<Z)
Line power:
W1=|VL|*|IL|*cos(30+(<Z))
W2=|VL|*|IL|*cos(30-(<Z))
In this case
W1= 380*220/10*cos(30+60)= 0
W2= 380*220/10*cos(30- 60)= 7240