If 3^(3k+4)+ 7^(2k+1) = 11a where a is an integer, then

3(3(k+1)+4)+ 7(2(k+1)+1) = 27×3(3k+4)+ 49×7(2k+1) = 27×11a + 22×7(2k+1)

and this is divisible by 11.

So, by the axiom of induction, 3⁽³ⁿ⁺⁴⁾+ 7⁽²ⁿ⁺¹⁾ is divisible by 11 for all positive integer values of n.