12=x+y+z

19=3*x+y

Note that x, y, z y and z are all integers:19 = 3 * 6+1;

So y= 1+3*n, n = 0, 1, 2. ...

n=0，y= 1，x=6，z = 5；

n= 1，y=4，x=5，z=3，

n=2，y=7，x=4，z= 1，

There are three situations.

The bonus is (1500 * x+700 * y) *11; I estimate that a team has only 1 1 players.