Let first watt meter be connected between R and B

$\begin{align*} \therefore \omega _1 &= V_{RB}.I_R \cos (V_{RB}-I_R)\\ &=V_{RB}.I_R \cos (30-\phi)\\ &=V_L I_L \cos (30-\phi)...(1)\\ \end{align*}$

($\because$ Voltage between any two terminals is line to line voltage or $V_L$)

Similarly, connecting between Y and B

$\begin{align*} \therefore \omega _2 &= V_{YB}.I_Y \cos (30+\phi)\\ &=V_L I_L \cos (30+\phi)...(2)\\ \end{align*}$

$\begin{align*} \therefore \omega_1 + \omega_2 &= V_L I_L [\cos (30-\phi)+\cos (30+\phi)]\\ &= V_L I_L [\cos 30.\cos \phi +\sin30. \sin \phi + \cos 30.\cos \phi -\sin 30.\sin \phi]\\ &=V_L I_L \left [ 2.\frac{\sqrt{3}}{2}.\cos \phi \right ]\\ &=\sqrt{3}V_LI_L \cos \phi \end{align*}$

but total $3-\phi$ power = $\sqrt{3}V_L I_L \cos \phi$

$\therefore \omega_1 +\omega_2$ = total power of a $3-\phi$ supply

Hence, $3-\phi$ power can be measured by 2 wattmeters.