I have the eigenvalue ${\cal{R}} _0$ of $P=\begin{pmatrix}0& \frac{\beta _h}{\mu +\gamma}&\frac{\beta _e}{\kappa (\mu +\gamma)}\\ 0& \frac{\beta _h\gamma}{(\mu +\gamma)(\mu +\alpha)} &\frac{\beta _e\gamma}{\kappa(\mu +\gamma)(\mu +\alpha)}\\ 0&\frac{\epsilon}{\delta} & 0\end{pmatrix}$ as
${\cal{R}} _0=\frac{\beta _h\gamma}{2(\mu+\gamma)(\mu+\alpha)}+\sqrt{\frac{\beta_h^2\gamma ^2}{4(\mu+\gamma)^2(\mu+\alpha)^2}+\frac{\beta _e\epsilon \gamma}{\kappa \delta (\mu+\gamma)(\mu+\alpha)}}$
To find the left eigenvector$(w_1,w_2,w_3)$ corresponding to ${\cal{R}} _0$, I solve
$(w_1,w_2,w_3)P={\cal{R}} _0(w_1,w_2,w_3)$
I get $w_1=0$ and I have a problem in finding $w_2$ and $w_3$ from the following equations
$\tag{1a}\frac{w_2\beta _h \lambda}{(\mu+\gamma)(\mu +\alpha)}+w_3\frac{\epsilon}{\delta}={\cal R}_0w_2$\ $\tag{1b} w_2\frac{\beta _e\gamma}{\kappa (\mu+\gamma)(\mu +\alpha) }={\cal R}_0w_3\label{6b}$
I was hoping to get the same expression for $w_3$ in both equations after substituting for ${\cal{R}} _0$ so that I can use a parameter to write the general solution, but I get
$w_3=\frac{-\beta_h\gamma \kappa \delta +\sqrt{(\beta_h\gamma \kappa \delta)^2+4\beta _e\epsilon \gamma\kappa \delta (\mu+\gamma)(\mu +\alpha)}}{2\epsilon (\mu+\gamma)(\mu +\alpha)\kappa }w_2$ and
$w_3= \frac{2\beta_e \gamma \delta}{\beta_h\gamma \kappa \delta +\sqrt{(\beta_h\gamma \kappa \delta)^2+4\beta _e\epsilon \gamma\kappa \delta (\mu+\gamma)(\mu +\alpha)}}w_2$