In a series LCR circuit, the total source voltage $V$ is related to the voltages across the resistor ($V_R$, read by $V_3$), inductor ($V_L$, read by $V_1$), and capacitor ($V_C$, read by $V_2$) by the phasor relationship: $V = \sqrt{V_R^2 + (V_L - V_C)^2}$ . Given that $V_1 = V_2 = 300$ V, the circuit is in a state of resonance where the inductive and capacitive voltages cancel each other out ($V_L - V_C = 0$). Therefore, $V = \sqrt{V_R^2} = V_R$. This implies the reading of voltmeter $V_3$ is equal to the source voltage. Based on the options and standard mains voltage conventions (and the specific data of this AIPMT 2010 problem where source is 220V, 50Hz, R=100\Omega ), the source voltage is 220 V. Thus, $V_3 = 220$ V. The current in the circuit at resonance is determined by the resistance: $I = V/R$. Using the probable answer values ($I = 2.2$ A) and derived voltage ($V = 220$ V), this is consistent with a resistance $R = 100 \, \Omega$.