Given three distinct points $x_1,x_2,x_3$ in $\mathbb F_2^n$ (endowed with the Hamming metric $d(\cdot,\cdot)$) with the same (but arbitrary) Hamming weight, the Chebyshev radius of them is defined as $$\min_{y\in\mathbb F_2^n}\max_{i\in\{1,2,3\}}d(x_i,y),$$ i.e., the radius of the smallest ball containing all of $x_1,x_2,x_3$. The Chebyshev centers of $x_1,x_2,x_3$ is the set of $y$'s which achieve the optimal value of the above optimization problem. Note that such $y$ is not necessarily unique. However, according to experiments, it seems that the set of Chebyshev centers always includes a $y$ such that $d(x_1,y)=d(x_2,y)=d(x_3,y)$. Let's call such $y$'s equidistant. Note that this is not true if we do not require constant weight. For instance, the Chebyshev center of $00,01,10$ is $00$, which is not equidistant. And $00$ happens to be the unique center in this case. My question is: how can one prove or disprove the above conjecture?
I ran a program for decent $n$'s, such as $n=7,8,9,10$ and did not find counterexamples. I tried to prove it using the method of types, known in information theory, but failed. I also tried proof by contradiction but also failed. Ideally we want to get a contradiction with the constant weight assumption, which is the only useful information we have.