$\frac{{\left({10}^2+{11}^2+{12}^2\right)}^{1000}}{{\left({13}^2+{14}^2\right)}^{1000}}=\frac{{\left(100+121+144\right)}^{1000}}{{\left(169+196\right)}^{1000}}=\frac{{365}^{1000}}{{365}^{1000}}={\left(\frac{365}{365}\right)}^{1000}=1^{1000}=1$