IMO D
There are 3 pairs of numbers -: a,x ; b,y ; c,z .
For the sum to be 1 as in the equation given , possible values of sqr(a-x) ,sqr(b-y) and sqr(c-z) could be 1,0,0 or 0,1,0 or 0,0,1.
For the time being lets suppose sqr(a-x) =1 and sqr(b-y) and sqr(c-z) are 0.
Possible values of a are 4,3,2,1 .
When a=4 ; y=3 -> 1 possible combination
when a=3 ; y = 2 or 4 -> so 2 possible combinations ....(as it is square of the difference(a-x) , the value is 1)
when a=2 ; y= 1 or 3 -> 2 possible combinations
when a=1 ; y= 2 or 0 -> 2 possible combinations
when a=0 ; y=1 -> 1 possible combination .
So we have a total 8 possible combinations for (a,x)
Now as (b-y) and (c-z) need to have value 0 , y and z will take the same value as b and c respectively .
What values can 'b' take -: 4,3,2,1,0 -> 5 values ; Corresponding to each of the 5 values of 'b' y will have only 1 value ( b and y need to be same to give sqr(b-y) = 0) -> total possible combinations -> 5*1 =5
What values can 'c' take -: 4,3,2,1,0 -> 5 values ; Corresponding to each of the 5 values of 'c' z will have only 1 value ( c and z need to be same to give sqr(c-z) = 0) -> total possible combinations -> 5*1=5
Total possible combinations = 8*5*5 = 200 .
Now in this example we chose (a-x) to have value 1 and (b-y) and (c-z) to have value 0 . Similarly , 2 other scenarios are also possible where (b-y) and (c-z) have values 1 . So total 3 different scenarios are possible.
so in total the number of possible non negative values of (a,x,b,y,c,z) are 200 * 3 = 600