The integer K = 3a^2 + 6b^2, where a and b are positive integers such

Let a = 2x, and b = 2y. Here x and y must have NO common factor since the greatest common factor of a/b, which is '2', is already taken outside. We need to keep this in mind that greatest common factor of x and y here will thus be 1 only. This also means that x and y both simultaneously CANNOT be divisible by 2.

So K becomes = 3*(2x)^2 + 6*(2y)^2 = 12*x^2 + 24*y^2 = 12*(x^2 + 2y^2).

Now 12 is already outside, so 12 has to be a factor of K. Inside the bracket we have x^2 + 2y^2. Since x and y both simultaneously cannot be divisible by 2, so its not necessary that x^2 + 2y^2 will be even, it could be odd also. This means the highest even factor of K, which we can be assured of, is '12'.

Hence D answer