Don't think you can directly say 6 = z - c, this is assuming that there is no carry over. It just so happens that you can't carry over on this question, but I believe you would have to do a check first to be sure.
I will try it in a different method... took me 2.5 mints
Problem states : All the 9 digits in the problem are different = no repeats.
(1) SUFFICIENT: Given 3a = f = 6y, the only possible value for y = 1. if y = 2, f> 9. Since y = 1, we know that f = 6 and a = 2.
now the question reads as
2 b c
+ d e 6
= x 1 z
using our knowledge of digits we can rewrite this as
200 + 10b + c + 100d + 10e + 6 = 100x + 10 + z
solving further ;
196 = 100(x - d) - 10(b + e) + 1(z - c)
our idea is to concentrate only on the units digit (z - c);
from the equation
6 = z - c.
lets plug in diffferent values;
z = 9 and c = 3
z = 8 and c = 2 ........ this cant be true as 'a' is already 2
z = 7 and c = 1 ........ this cant be true as 'y' is already 1.
hence Z =9 and X = 3... perfect.
(2) INSUFFICIENT: The statement f - c = 3 .. provides us with different values... not possible.
Hence A.