I am not sure that I understand your requirements - and am confused by the three simultaneous formats. Do you mean that you have three different ways of scoring the same matches?
If I have understood correctly that the twosomes are different on each day and they go around the course 2 twosomes at a time, then the following might help:
(11 9 v 6 7) (1 2 v 12 8) (10 5 v 4 3)
(12 7 v 4 8) (2 3 v 10 9) (11 6 v 5 1)
(10 8 v 5 9) (3 1 v 11 7) (12 4 v 6 2)
(10 6 v 7 8) (4 1 v 3 9) ( 2 12 v 5 11)
(11 4 v 8 9) (5 2 v 1 7) ( 3 10 v 6 12)
(12 5 v 9 7) (6 3 v 2 8) ( 1 11 v 4 10)
That's only 6 rounds so you would have to make up a 7th using twosomes that have not yet occured.
I have something similar for 12 rounds:
( 2 3 v 12 6) (10 11 v 5 9) ( 4 1 v 8 7)
( 3 1 v 10 4) (11 12 v 6 7) ( 5 2 v 9 8)
( 1 2 v 11 5) (12 10 v 4 8) ( 6 3 v 7 9)
(10 6 v 12 2) ( 1 5 v 7 3) ( 9 11 v 4 8)
(11 4 v 10 3) ( 2 6 v 8 1) ( 7 12 v 5 9)
(12 5 v 11 1) ( 3 4 v 9 2) ( 8 10 v 6 7)
( 6 8 v 9 1) ( 5 3 v 10 2) (11 7 v 4 12)
( 4 9 v 7 2) ( 6 1 v 11 3) (12 8 v 5 10)
( 5 7 v 8 3) ( 4 2 v 12 1) (10 9 v 6 11)
( 4 7 v 6 5) ( 2 8 v 3 11) (12 9 v 10 1)
( 5 8 v 4 6) ( 3 9 v 1 12) (10 7 v 11 2)
( 6 9 v 5 4) ( 1 7 v 2 10) (11 8 v 12 3)
All possible twosomes occur (either on the left or right of the 'v') at least once, six occur twice (1&12 2&10 3&11 6&7 8&4 9&5).