$V_{R}=V_B =V_Y=V_{ph}$
$V_{L}=V_{RB}=V_{BY}=V_{RY}$
$\begin{align*}
\therefore V_L &=\bar{V_R} - \bar{V_B}\
&=V_R+j0-V_B\cos 60 -jV_B \sin 60\
&= V_{ph} -V_{ph}\cos 60 -jV_{ph} \sin 60\
&=V_{ph}(1-\cos 60)-jV_{ph}\sin 60\
&=V_{ph}\sqrt{(1-\cos 60)^2 +(\sin 60)^2} =1.732V_{ph} =\sqrt{3}V_{ph}
\end{align*}
$
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