Y_{S1}\delta_{2})^2Y^2_{3}v^2}{2M_{2}(4M^2_{2} - Y^2_{S1}\delta^2_{2})}  \;\; , \end{aligned}$$ and $$\begin{aligned}
\label{numassp3}
m_{\nu_{3}} &\approx& \frac{3Y^2_{3}v^2}{M_{2}} + \frac{9Y^2_{3}v^2}{2M_{2} - Y_{S1}\delta_{3}} \;\; ,  \nonumber \\
m_{\nu_{2}} &\approx& -\frac{3Y^2_{3}v^2}{2M_{2}} + \frac{2Y^2_{3}v^2}{2M_{2} + Y_{S1}\delta_{3}} + \frac{Y^2_{3}v^2(2M_{2} + Y_{S1}\delta_{3})^2}{8M^3_{2}} \;\; ,  \nonumber \\
m_{\nu_{1}} &\approx& \frac{6Y^2_{3}v^2}{M_{2}} + \frac{Y^2_{3}v^2}{2M_{2} + Y_{S1}\delta_{3}} + \frac{3Y^2_{3}v^2(2M_{2} + Y_{S1}\delta_{3})}{4M^2_{2}} + \frac{Y^2_{3}v^2(2M_{2} + Y_{S1}\delta_{3})^2}{8M^3_{2}}  \;\; , \end{aligned}$$ respectively. Note that the above three sets of equations reduce to Eq.(\[neutrinosmass0\]) when $\delta_i \rightarrow 0$ as they should. In these three cases we have the decoupling fourth neutrino mass which should be required to satisfy $$\begin{aligned}
m_{\nu_4} &=& \frac{Y^2_{1}v^2_{H_1}}{M_1} + (\frac{2}{2M_{2} + Y_{S1}\delta_{i}} + \frac{1}{2M_{2}})Y^2_{3}v^2 > M_{Z}/2 \;\; .\end{aligned}$$ By assuming the new physics to be around $\sim 1$ TeV, and taking $v_{H_1} = 220$ GeV, $v = 10$ GeV, $Y_{S1}\delta_{i=1,2,3} = 400$ GeV, $M_{1} = 500$ GeV, $M_{2} = 10^8$ GeV, $Y_{1} = 1$, and $Y_{3} = 10^{-3}$, we have $$\begin{aligned}
\label{nu-mass}
m_{\nu_3} &\approx& 0.75\times10^{-2}~{\rm eV} \;\;, \quad m_{\nu_{2}} \approx 5.4\times10^{-6}~{\rm eV} \;\;, \quad m_{\nu_{1}} \approx 0.35\times10^{-2}~{\rm eV}\; \; ; 
\nonumber \\
m_{\nu_{3}} &\approx& 0.75\times10^{-2}~{\rm eV} \;\;, \quad m_{\nu_{2}} \approx 6.0\times10^{-9}~{\rm eV} \;\;,  \quad m_{\nu_{1}} \approx 0.70\times10^{-2}~{\rm eV} \;\; ; 
\nonumber \\
m_{\nu_{3}} &\approx& 0.75\times10^{-2}~{\rm eV} \;\;, \quad m_{\nu_{2}} \approx 6.0\times10^{-15}~{\rm eV} \;\;, \quad m_{\nu_{1}} 