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\begin{align}
\pars{a\,\sigma_{0} + \vec{r}\cdot\vec{\sigma}}\ket{\Psi} = E\ket{\Psi}
& \implies\quad
\vec{r}\cdot\vec{\sigma}\ket{\Psi} = \pars{E - a}\ket{\Psi}
\label{1}\tag{1}
\\[5mm] \implies
\pars{\vec{r}\cdot\vec{\sigma}}^{2}\ket{\Psi} =
\pars{E - a}\vec{r}\cdot\vec{\sigma}\ket{\Psi}
& \implies\quad
\vec{r}\cdot\vec{\sigma}\ket{\Psi} = {r^{2} \over E-a}\ket{\Psi}
\label{2}\tag{2}
\end{align}
Compare \eqref{1} and \eqref{2}:
$$
E - a = {r^{2} \over E - a}\quad\implies\quad
\bbox[#ffb,5px,border:2px groove navy]{\ E = a \pm r = a \pm\verts{\vec{r}}\ }
$$
Note that
$$
\pars{\vec{r}\cdot\vec{\sigma}}^{2} =
\vec{r}\cdot\vec{\sigma}\,\,\vec{r}\cdot\vec{\sigma} =
\vec{r}\cdot\vec{r} +
\ic\,\vec{\sigma}\cdot\underbrace{\vec{r} \times \vec{r}}_{\ds{=\ \vec{0}}} =
r^{2}
$$