It will not be possible to find a perfect solution.
With 11 players there are 11*10/2 or 55 possible pairings. You wish them to play 3 times, so a total of 165 pairings, but unfortunately a number that is not divisible by 6, the number of pairings per round.
Cyclic schedules for multiples of 11 rounds may be suitable as this has the advantage of giving every player the same number of matches. For example this plan
1 : 5 6 10 1
2 : 6 7 11 2
3 : 7 8 1 3
4 : 8 9 2 4
5 : 9 10 3 5
6 : 10 11 4 6
7 : 11 1 5 7
8 : 1 2 6 8
9 : 2 3 7 9
10 : 3 4 8 10
11 : 4 5 9 11
12 : 5 2 10 11
13 : 6 3 11 1
14 : 7 4 1 2
15 : 8 5 2 3
16 : 9 6 3 4
17 : 10 7 4 5
18 : 11 8 5 6
19 : 1 9 6 7
20 : 2 10 7 8
21 : 3 11 8 9
22 : 4 1 9 10
23 : 5 2 6 3
24 : 6 3 7 4
25 : 7 4 8 5
26 : 8 5 9 6
27 : 9 6 10 7
28 : 10 7 11 8
29 : 11 8 1 9
30 : 1 9 2 10
31 : 2 10 3 11
32 : 3 11 4 1
33 : 4 1 5 2
gives 22 of the 55 pairs exactly 3 matches, while the remaining 33 pairs get 4 matches.
While this plan:
1 : 11 1 2 8
2 : 1 2 3 9
3 : 2 3 4 10
4 : 3 4 5 11
5 : 4 5 6 1
6 : 5 6 7 2
7 : 6 7 8 3
8 : 7 8 9 4
9 : 8 9 10 5
10 : 9 10 11 6
11 : 10 11 1 7
12 : 7 3 9 1
13 : 8 4 10 2
14 : 9 5 11 3
15 : 10 6 1 4
16 : 11 7 2 5
17 : 1 8 3 6
18 : 2 9 4 7
19 : 3 10 5 8
20 : 4 11 6 9
21 : 5 1 7 10
22 : 6 2 8 11
gives 33 of the pairs exactly 2 matches, while the remaining 22 pairs get 3 matches.
Hope that helps.