if a b c are three natural numbers in Arithmetic progression and a+b+c=21 then all the possible values of the ordered triplet (a,b,c) is
Your AP is $\displaystyle n,~n+d,~n+2d$ so $\displaystyle n+n+d+n+2d=3n+3d=21$.
That means that $\displaystyle n+d=7$.
The numbers of pairs that satisfy that depends on if we consider 0 a natural or not.
Here is an example: $\displaystyle n=3~\&~d=4$ then the AP is $\displaystyle 3,~7,~11$.
Can you finish?