Two generators 200∠30° and 210∠50° V

The given terminal voltages are taken as per-phase (line-to-neutral) values, since the generators are star connected with neutral grounded and the impedance is per phase. Three-phase power is 3 times the per-phase power.

Let I flow from G1 to G2 through $Z = 2 + j8\ \Omega$. Then $S_1 = V_1I^*$ is the power delivered by G1 and $S_2 = V_2I^*$ is the power received by G2.

Current

$$\begin{aligned} V_1 &= 200\angle30^\circ = 173.21 + j100.00\ \text{V}\\ V_2 &= 210\angle50^\circ = 134.99 + j160.87\ \text{V}\\ I &= \frac{V_1 - V_2}{Z} = \frac{38.22 - j60.87}{2 + j8}\\ &= -6.037 - j6.287 = 8.716\angle-133.84^\circ\ \text{A} \end{aligned}$$

Powers per phase

$$\begin{aligned} S_1 &= V_1I^* = (200\angle30^\circ)(8.716\angle133.84^\circ) = -1674.31 + j485.19\ \text{VA}\\ S_2 &= V_2I^* = (210\angle50^\circ)(8.716\angle133.84^\circ) = -1826.24 - j122.55\ \text{VA}\\ S_{line} &= |I|^2Z = 75.97(2 + j8) = 151.94 + j607.74\ \text{VA} \end{aligned}$$

Interpretation

Check: $1826.24 - 1674.31 = 151.93$ W and $485.19 + 122.55 = 607.74$ var.

Real power flows from G2 to G1 because $V_2$ leads $V_1$ by 20°. G1 behaves as a synchronous motor. Both machines supply reactive power, which is consumed by the line reactance.

Answer: G2 generates 5.48 kW and 0.37 kvar. G1 absorbs 5.02 kW (motor) and generates 1.46 kvar. The line absorbs 0.46 kW and 1.82 kvar (three-phase totals).