Explanation

N ≡ 5 modulo 6, 12, 15 and 18 means N−5 is a common multiple of all four, so N−5 is a multiple of lcm(6,12,15,18) = 180. Write N = 180m+5. The multiple-of-11 condition gives 180m+5 ≡ 0 (mod 11), i.e. 4m ≡ 6 (mod 11), so m ≡ 7 (mod 11). The least positive m is 7, giving N = 1265, which exceeds 1200 — option D. (The stored option labels appear truncated from range-form options.)