1 & 0 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\0 & \sqrt{\frac{1}{3}} & \sqrt{\frac{1}{3}} & \sqrt{\frac{1}{3}} \\0 & -\sqrt{\frac{1}{6}} & -\sqrt{\frac{1}{6}} & \sqrt{\frac{2}{3}} \end{array}\right).\end{aligned}$$ There are six conditions corresponding to the vanishing off diagonal terms, viz. $$\begin{aligned}
\label{con1}
&&E = H \quad {\rm ,} \quad G = I \;\; ,\\
\label{con2}
&&G + J = E + F \;\; , \\
&&B = C = D = 0 \;\; .
\label{con3}\end{aligned}$$ Eq. (\[con1\]) implies $v_{\phi_1} = v_{\phi_2}$ and the solution for Eq. (\[con2\]) is $$\begin{aligned}
Y_{3}(v_{\phi_1} - v_{\phi_3})\left[Y_{3}(2v_{\phi_1} + 3v_{\phi_3}) -2Y_{2}v_{H_1}\right]
& = & 0 \; . \end{aligned}$$ Together with the conditions from Eq. (\[con3\]) we have only one consistent solution, $v_{\phi_1} = v_{\phi_2}$ = $v_{\phi_3} \equiv v$. Thus we write $$\begin{aligned}
\label{vacuum} 
\langle \Phi_{3} \rangle= (v,v,v) \end{aligned}$$ and the constraint from Eq. (\[con3\]) becomes $$\begin{aligned}
v_{H_1} = -\frac{2Y_{3}}{Y_{2}}v \;\;.\end{aligned}$$ This leads to the four neutrino masses $$\begin{aligned}
m_{\nu_4} &=& \frac{Y^2_{1}v^2_{H_1}}{M_1} + \frac{3Y^2_{3}v^2}{2M_{2}} \;\; , \nonumber \\
m_{\nu_3} &=& \frac{15Y^2_{3}v^2}{2M_{2}} \;\; , \nonumber \\
m_{\nu_2} &=& 0 \;\; , \nonumber \\
m_{\nu_1} &=& m_{\nu_3} \; \; .
\label{neutrinosmass0}\end{aligned}$$ The parameters in the first term of $m_{\nu_4}$ all differ from those in the expressions for 