Start with:
PQRSTU | PQRSTU | PQRSTU | PQRSTU | PQRSTU | PQRSTU LEFT }}
ABCDEF | ABCDEF | ABCDEF | ABCDEF | ABCDEF | ABCDEF {{ LEFT
1) B is to the immediate left of C, who does not face Q. Q is third to the left of R.
2) Neither Q nor C sit at the ends. B is to the right of F and D.
2 means we can easily remove Q and C. B cannot be in 1 or 2 (right of F and D), or 6 (left of C). F and D cannot be in 6. B cannot be in 5 (there is no C in 6). C cannot be in 2 or 3 (no B in 1 or 2). F and D can be removed from 5 (no B to the right).
P RSTU | PQRSTU | PQRSTU | PQRSTU | PQRSTU | P RSTU LEFT }}
A DEF | A DEF | AB DEF | ABCDEF | A C E | A E {{ LEFT
3) D faces the person who is third to the right of T, who faces E.
T cannot be 1,2,3 (there is someone 3 to the right), E is 4,5,6.
From 1) Q is 4,5,6, R is 1,2 (not 3 as no Q in 6)..
P RS U | P RS U | P S U | PQ STU | PQ STU | P STU LEFT }}
A D F | A D F | AB D F | ABCDEF | A C E | A E {{ LEFT
4) P sits second to the left of the person who faces A.
Tells us A is not 5,6, P is not 1,2. E fixed, so we can remove it from the others, fixing C, so remove C too. B is 4 (from 1).
3) tells us T is facing E, so T fixed.
4) says A not in 4. 3) says D is 3, and so 4) says P not 5.
RS U | RS U | P S U | PQ S U | Q S U | T LEFT }}
A F | A F | D | B | C | E {{ LEFT
5) S is not an immediate neighbor of P.
S is not 3,4.
1) says Q not 5, therefore Q is 4, and R is fixed too.
SU lock in (2,5) says P is 3.
R | S U | P | Q | S U | T LEFT }}
A F | A F | D | B | C | E {{ LEFT
4) gives A is 1. F is 2. 5 says S is not 2.
R | U | P | Q | S | T LEFT }}
A | F | D | B | C | E {{ LEFT
So:
RUPQST
AFDBCE