Read the given information carefully and answer the questions that follow.…

In linear (single-row) seating puzzles, every relative-position clue — immediate left/right, nth to the left/right, at an end — fixes one person's seat relative to another. Anchor the arrangement at whichever person is confirmed to be at an end (an absolute position), then chain the remaining clues outward from that anchor until all seats are assigned.

Applying this to the given clues:

1. Q is at one of the two ends, so take Q's seat as the anchor at position 1. Since M is second to the right of Q, M is at position 3.

2. V is the immediate neighbour of both M and N, so V sits directly between them. With M at position 3, V must be at position 4, and N is at position 5 on V's other side.

3. The clue states V is third to the left of S, meaning S sits three seats to the right of V. With V at position 4, S is at position 4 + 3 = 7.

4. T is third to the right of U, T is immediate left of P, and R is immediate right of P; also P is second to the right of O. So T, P and R occupy three consecutive seats, with O two seats before P and U three seats before T.

5. Among the seats not yet assigned (2, 6, 8, 9, 10 and 11), two runs of three consecutive seats are possible: 8-9-10 and 9-10-11. Testing 8-9-10 gives O = P − 2 = 7, but seat 7 is already taken by S — a contradiction. So the valid run is 9-10-11: T = 9, P = 10, R = 11.

6. Then O = P − 2 = 8, and U = T − 3 = 6. The one seat left over, position 2, goes to W, the only friend not yet placed.

Cross-check against the final order Q(1) W(2) M(3) V(4) N(5) U(6) S(7) O(8) T(9) P(10) R(11): T is immediately left of P and three right of U; V is between M and N and three left of S; M is two right of Q, who is at an end; R is immediately right of P, who is two right of O — every condition holds.

The row has 11 seats in total, so the centre seat is position 6 — occupied by U.