Also you have not mentioned n which I assume is 3

n= 3 a=4 b=7 c=13 is a good example however

Despite this, if you also add the conditions that a, b, c are positive integers and also that
mod(a,n)= mod(b plus c, n)
mod(b,n)= mod(a plus c, n)
mod(c,n)= mod(a plus b, n)
You have an impossible situation.

This is a straightforward idea followed up shortly.

>10 + 11 + 12 = 23
>
>In general, for a prime n = 2k + 3 take
>
>a = k, b = k+1, c = k+2.

This is nonsense because 2 ≠ 3.

But then, e.g., take

n = 11, a = 2, b = 3, c = 6, or
n = 13, a = 3, b = 4, c = 6, or
n = 13, a = 2, b = 5, c = 6, or
n = 17, a = 3, b = 5, c = 9, or ...

So, it is too easy. You probably had something else in mind.

13 = 1+3+9 = 1+5+7
17 = 1+3+13 = 1+5+11 = 1+7+9 = 3+5+9

...and so on and so fourth.

As far as adding the mod restriction, it is indeed impossible (as by definition each of a, b, and c would have to be some integer multiple of n).