Maximum power over 20 + j50 Ω line

Short line: $Z = 20 + j50 = 53.85\angle68.20^\circ\ \Omega$, $V_S$ = 33 kV, $V_R$ = 31.2 kV (line values, so powers are three-phase).

Maximum active power

For a short line, $P_R$ is maximum when $\delta = \theta$:

$$\begin{aligned} P_{R,max} &= \frac{|V_S||V_R|}{|Z|} - \frac{|V_R|^2R}{|Z|^2} \\ &= \frac{33\times31.2}{53.85} - \frac{31.2^2\times20}{53.85^2} \\ &= 19.12 - 6.71 = 12.41\ \text{MW} \end{aligned}$$

Receiving-end power factor

At $\delta = \theta$:

$$Q_R = -\frac{|V_R|^2X}{|Z|^2} = -\frac{31.2^2\times50}{2900} = -16.78\ \text{MVAr}$$ $$\text{pf}_R = \frac{12.41}{\sqrt{12.41^2 + 16.78^2}} = 0.594\ \text{leading}$$

(Negative $Q_R$: the load must supply reactive power, i.e. it is leading.)

Line efficiency

$$\begin{aligned} I &= \frac{\sqrt{12.41^2 + 16.78^2}}{\sqrt3\times31.2} = 0.3862\ \text{kA} = 386.2\ \text{A} \\ P_{loss} &= 3I^2R = 3(386.2)^2(20) = 8.95\ \text{MW} \\ \eta &= \frac{12.41}{12.41 + 8.95}\times100 = 58.1\% \end{aligned}$$

(Check: the sending-end power $P_S = \frac{|V_S|^2R}{|Z|^2} - \frac{|V_S||V_R|}{|Z|}\cos 2\theta = 7.51 + 13.84 = 21.35$ MW, so the loss is $21.35 - 12.41 = 8.94$ MW.)

Answer: $P_{max}$ ≈ 12.41 MW, receiving-end pf ≈ 0.594 leading, efficiency ≈ 58.1 %.