Follow the 'schedules' link near the top of this page then enter 12 items and 'Whist'. Treat each table of 4 players as a foursome and this will solve the general 12 player problem. (Randomize the tee-off order so that one player is not always in the first group of 4.)
The scenario where two of the players are father and son, can never be fully balanced, but I think the schedule below comes as close as possible.
(B J A D) (H I C G) (E F Y Z)
(F H D J) (C E A G) (B I Y Z)
(B F G I) (A E C D) (H J Y Z)
(B E H J) (D I A F) (C G Y Z)
(C F G J) (E H B I) (A D Y Z)
(B D C I) (G H A J) (E F Y Z)
(C H D F) (A I E J) (B G Y Z)
(E G D H) (B C F J) (A I Y Z)
(F I A H) (D E B G) (C J Y Z)
(C J E I) (F G A B) (D H Y Z)
(D G I J) (B H A C) (E F Y Z)
The father and son are Y and Z and will need to get on well with players E and F as they play three times in the same foursome. Players A to D and G to J all play together either three or four times.
