We know that 14 (A2 + B2 + C2) = (a + 2B + 3C) 2, and prove that a: B: C = 1:2:3

It is proved that: ∵ 14 (A2 + B2 + C2) = (a + 2B + 3C) 2, ∵ 14a2 + 14b2 + 14C2 = A2 + 4B2 + 9c2 + 4AB + 6ac + 12bc, ∵ 4a2-4ab + B2 + 9a2-6ac + C2 + 9b2-12bc + 4C2 = 0, ∵ 2a-b) 2 + (3a-c) 2 + (3b-2c) 2 = 0, ∵ 2a-b = 0, 3a-c = 0, 3b-2c = 0, ∵ B = 2A, C = 3A, a: B: C = 1:2:3