For any number, it is divisible by 9 if all of the individual numbers in the number are divisible by 9 when added together.
For example:
18 (9x2): 1+8 = 9/9 = 1,
36 (9x4): 3+6 = 9/9 = 1
396 (9x44): 3+6+9 = 18/9 = 2
657 (9x73): 6+5+7 = 18/9 = 2
999 (9x111): 9+9+9 = 27/9 = 3
etc.
The remainder when the individual numbers in a number are divided by 9 is the remainder when the number is divided by 9 as well
For example:
19 (9x2 r1): 1+9 = 10/9 = 1 r1
38 (9x4 r2): 3+8 =11/9 = 1 r2
397 (9x44 r1): 3+9+7 = 19/9 = 2 r1
659 (9x73 r2): 6+5+9 = 20/9 = 2 r2
1002 (9x111 r3): 1+0+0+2 = 3/9 = 0 r3
etc.
That being said, in the problem if x, y, and z are all different positive integers less than 4, they can be 1,2, and 3 in any order:
1230, 1320, 2130, 2310, 3120, and 3210
if we apply the rule above:
1230: 1+2+3+0 = 6/9 = 0 r6
1320: 1+3+2+0 = 6/9 = 0 r6
2130: 2+1+3+0 = 6/9 = 0 r6
etc. etc. etc.
If you know this rule, then all you would need to know is x+y+z = 1+2+3 = 6
Answer is C