Step 1: Since Q is at one of the ends and M is the second to the right of Q, Q cannot be at the right end (there would be no room for M). So Q is at position 1 and M is at position 3.
Step 2: From clue (D), R is one place to the right of P, so R = P + 1, and P is second to the right of O, so O = P minus 2. From clue (A), T is one place to the left of P, so T = P minus 1, and T is third to the right of U, so U = P minus 4. This gives a fixed block: U, _, O, T, P, R at positions P-4, P-3, P-2, P-1, P, P+1.
Step 3: Testing values of P from 5 to 10 so all eleven positions stay within 1 to 11 and none clash with Q (position 1) or M (position 3), only P = 10 works without a clash: U=6, O=8, T=9, P=10, R=11.
Step 4: From clue (B), V must be an immediate neighbour of both M and N, so V is at position 2 or 4. Position 2 would force N onto position 1, already taken by Q, so V = 4 and N = 5. Then S = V + 3 = 7, which is free. The one remaining position, 2, goes to W.
Step 5: Final row (positions 1 to 11): Q, W, M, V, N, U, S, O, T, P, R. The center position of 11 people is position 6, which is U.