A  1-rotational (125,5,1)-RBIBD (over Z_2 x Z_2 x Z_31).

The base blocks are 
   B_1 = {(0,0,1), (0,0,2), (0,0,4), (0,0,8), (0,0,16)};
   B_2 = {(0,0,3), (0,0,29), (0,1,20), (1,0,25), (1,1,26)};
   B_3 = {(0,0,6), (0,0,27), (0,1,9), (1,0,19), (1,1,21)};
   B_4 = {(0,0,12), (0,0,23), (0,1,18), (1,0,7), (1,1,11)};
   B_5 = {(0,0,24), (0,0,15), (0,1,5), (1,0,14), (1,1,22)};
   B_6 = {(0,0,17), (0,0,30), (0,1,10), (1,0,28), (1,1,13)};
   B_7 = {(0,0,0), (0,1,0), (1,0,0), (1,1,0),$\infty$}.

The above design cannot be isomorphic to AG(3,5) because it possesses
Z_2 x Z_2 as an automorphism group fixing one point. On the other hand
it is easy to see that the full stabilizer of a point in AG(3,5) does
not have Z_2 x Z_2 as a subgroup.