Solution:

By the power rule for exponents, $$a^{(\log_3 7)^2} = \left(a^{\log_3 7}\right)^{\log_3 7} = 27^{\log_3 7} = \left(3^{\log_3 7}\right)^3 = 7^3 = 343.$$

In the same way, $b^{(\log_7 11)^2} = 49^{\log_7 11} = \left(7^{\log_7 11}\right)^2 = 11^2 = 121,$ and $c^{(\log_{11} 25)^2} = \left(\sqrt{11}\right)^{\log_{11} 25} = \left(11^{\log_{11} 25}\right)^{1/2} = 25^{1/2} = 5.$

The sum is $343 + 121 + 5 = 469.$