To solve this problem, we need to analyze the given lossless power network with three buses. The key information is that the voltage magnitudes at all buses are 1 per unit (pu), and the differences in their phase angles are small. The real power flow from bus 1 to bus 2 is 0 pu.

Given that \( P_1 = mP_2 \) and the power flow \( P_{12} = 0 \), we need to determine the relationship between the reactances to satisfy these conditions.

From power flow theory, the power flow between two buses \( i \) and \( j \) in terms of voltages and reactance is given by:

P_{ij} = \frac{V_i V_j}{X_{ij}} \sin(\theta_i - \theta_j)

Since the power flow from bus 1 to bus 2 is 0, we have:

P_{12} = \frac{V_1 V_2}{j\alpha} \sin(\theta_1 - \theta_2) = 0

For \( P_{12} = 0 \), \( \sin(\theta_1 - \theta_2) = 0 \), indicating that \( \theta_1 = \theta_2 \) or very close. Hence, we focus on the relationship between the injections \( P_1 \) and \( P_2 \).

Since \( P_1 = mP_2 \), examine the power balance equation at each bus using Kirchhoff’s Current Law:

• At bus 1: \( P_1 = (\beta + \alpha) \sin(\theta_1 - \theta_3) \)
• At bus 2: \( P_2 = (\gamma + \alpha) \sin(\theta_2 - \theta_3) \)

Since both angles \( \theta_1 \approx \theta_2 \) due to \( P_{12} = 0 \), \( \sin(\theta_1 - \theta_3) \approx \sin(\theta_2 - \theta_3) \). The ratio of \( P_1 \) and \( P_2 \) becomes:

\frac{P_1}{P_2} = \frac{(\beta + \alpha) \sin(\theta_1 - \theta_3)}{(\gamma + \alpha) \sin(\theta_2 - \theta_3)}

\Rightarrow \frac{(\beta + \alpha)}{(\gamma + \alpha)} = m

This needs to simplify while maintaining differentials. Given slight angles and equal contributions, a variable separation leads us to the condition that:

\gamma = m\beta,

Therefore, the correct option is \(\gamma = m\beta\).