Given ellipse is vertical $$ \begin{aligned} & \therefore \mathrm{f}(3 \mathrm{a}+15)>\mathrm{f}\left(\mathrm{a}^2+7 \mathrm{a}+3\right) \\ & \because \mathrm{f}(\mathrm{x}) \text { is decreasing } \forall \mathrm{x} \in \mathrm{R} \\ & \Rightarrow 3 \mathrm{a}+15<\mathrm{a}^2+7 \mathrm{a}+3 \\ & \Rightarrow \mathrm{a}^2+4 \mathrm{a}-12>0 \end{aligned} $$
$\begin{aligned} & \Rightarrow(\mathrm{a}+2)^2>16 \\ & \Rightarrow|(\mathrm{a}+2)|>4 \\ & \Rightarrow \mathrm{a}>2 \text { or } \mathrm{a}<-6 \\ & \Rightarrow \mathrm{a} \in \mathrm{R}-[-6,2] \\ & \therefore \alpha=-6, \beta=2 \\ & \therefore \alpha^2+\beta^2=36+4=40\end{aligned}$