A curious phenomenon in Number Theory (related to Algebraic Geometry)

Let us consider a prime number $p$ and three distinct positive integers $n_1<n_2<n_3$ less than $p$.

Let us assume that the triple $(n_1, n_2, n_3)$ satisfies the following condition

$$k+[kn_1]+[kn_2]+[kn_3]=2p, \ \ for \ all \ 1\leq k\leq p-1$$

Where $[kn_i]$ denotes the rest of the division of $kn_i$ by $p$.

For example, the triple $(n, p-n, p-1)$ satisfies my condition. Indeed, $[k(p-n)]=p-[kn]$ and so for all $k$ we have

$$k+[kn]+(p-[kn])+p-k=2p.$$

My question is the following: Is it true that the only possible triples satisfying my condition are $(n,p-n,p-1)$?

As an example, for $p=11$, if you choose the triple $(4, 8, 9)$, you would get

$$1+4+8+9=22 \ for \ k=1$$

$$2+8+5+7=22 \ for \ k=2$$

$$3+1+2+5=11 \ for \ k=3$$

Since the condition is not satisfied for $k=3$, then that triple is discarded. Indeed one can simple see that for $p=11$ the only admissible triple is $(n, 11-n, 10)$ for $2\leq n\leq 5$.

I think this is a very deep result. I met this problem during my research, by studying the rationality of $\mathbb Z/p$-Galois coverings of the projective plane $\mathbb P^2$ with a simple normal crossing branch locus and with geometric genus $p_g(S)=0$.