Let a = |x|,b = |y|,c = |z|
Therefor we need to find the numbers of ways we can select 3 numbers a,b,c (a>0,b>0,c>0) such that
a+b+c=12
12 = 1+1+1+1+1+1+1+1+1+1+1+1
Whenever we select 2 "+" signs and replace them with a comma we get 3 integers p,q,r whose sum is 12.
Number of ways of selecting the 2 "+" signs from 11 ones is:
11C2=55