$$\begin{aligned} &\begin{aligned} \Delta & =\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 10 \end{array}\right|=1(20-16)-1(10-4)+1(4-2) \\ & =4-6+2=0 \end{aligned}\\ &\text { For infinite solutions }\\ &\begin{aligned} & \Delta_{\mathrm{x}}=\Delta_{\mathrm{y}}=\Delta_{\mathrm{z}}=0 \\ & \mathrm{~m}^2-3 \mathrm{x}+2=0 \\ & \mathrm{~m}=1,2 \\ & \alpha=1, \beta=2 \\ & \therefore \sum_{\mathrm{n}=1}^{10}\left(\mathrm{n}^\alpha+\mathrm{n}^\beta\right)=\sum_{\mathrm{n}=1}^{10} \mathrm{n}^1+\sum_{\mathrm{n}=1}^{10} \mathrm{n}^2 \\ &= \frac{10(11)}{2}+\frac{10(11)(21)}{6} \\ &= 55+385 \\ &= 440 \end{aligned} \end{aligned}$$