The correct option is A 224
If t=0:x+y+z=15. This has 15+3−1C3−1=136 solutions.
If t=1:x+y+z=10. This has 10+3−1C3−1=66 solutions.
If t=2:x+y+z=5. This has 5+3−1C3−1=21 solutions.
If t=3:x+y+z=0. This has only 1 solution.
Thus, answer =136+66+21+1=224
Hence, (B) is correct.