The sum of three consecutive multiples of 11 is 363. What is the largest of these three multiples?

Let the three consecutive multiples of 11 be 11n, 11(n+1), 11(n+2). Their sum is 11(3n + 3) = 33(n + 1). Setting 33(n+1) = 363 gives n + 1 = 11, so n = 10. The three multiples are 110, 121, 132, and the largest is 132. Distractor 143 belongs to the next set (121+132+143 = 396), and 121 is the middle multiple, not the largest.