Two positive consecutive multiples of 3 can be written as 3*k and 3*(k+1), where k is positive integer

(3*k)^2 + (3*(k+1))^2 = 369

9*k^2 + 9*k^2 + 18*k + 9 = 369

18*k^2 + 18*k - 360 = 0

k^2 + k - 20 = 0

By solving this quadratic equation we get

k = - 5 is not a solution since k needs to be positive

k = 4 is a solution

Indeed 12^2 + 15^2 = 369.

The two consecutive multiples of 3 are 12 and 15.