I had some inspiration and think the solution below will work well. From the perspective of one team, all 8 normal volleyball opponents are unique, and all 6 tri-ball opponents are unique. Of course there are only 10 possible opponents so there must be overlap, therefore each team will oppose 4 of the other teams twice, once as a normal opponent and once as a tri-ball opponent.
( 4 6) ( 8 11) (10 5) ( 2 1) ( 3 7 9)
( 5 7) ( 9 1) (11 6) ( 3 2) ( 4 8 10)
( 6 8) (10 2) ( 1 7) ( 4 3) ( 5 9 11)
( 7 9) (11 3) ( 2 8) ( 5 4) ( 6 10 1)
( 8 10) ( 1 4) ( 3 9) ( 6 5) ( 7 11 2)
( 9 11) ( 2 5) ( 4 10) ( 7 6) ( 8 1 3)
(10 1) ( 3 6) ( 5 11) ( 8 7) ( 9 2 4)
(11 2) ( 4 7) ( 6 1) ( 9 8) (10 3 5)
( 1 3) ( 5 8) ( 7 2) (10 9) (11 4 6)
( 2 4) ( 6 9) ( 8 3) (11 10) ( 1 5 7)
( 3 5) ( 7 10) ( 9 4) ( 1 11) ( 2 6 8)