Let $\mathbf a, \mathbf b, \mathbf c$ be three vectors. We want to minimize the function $$ f(\mathbf r)=|\mathbf r-\mathbf a|+|\mathbf r-\mathbf b|+|\mathbf r-\mathbf c|. $$ Now as usual we evaluate the gradient: $$ \nabla f=\frac{\mathbf r-\mathbf a}{|\mathbf r-\mathbf a|}+\frac{\mathbf r-\mathbf b}{|\mathbf r-\mathbf b|}+\frac{\mathbf r-\mathbf c}{|\mathbf r-\mathbf c|}=0. $$ The sum of three unit vectors is zero, so the angle between any two of them is $\frac{2}{3}\pi$.
Question: can I find an expression of $\mathbf r$ in terms of $\mathbf a, \mathbf b, \mathbf c$? Please avoid writing the vectors into their components. I am looking forward to seeing a beautiful solution.
A beautiful solution can indeed be derived, with all the symmetry preserved. It is suitably more involved, as expected.
Start with the zero gradient condition
$$ \frac{\mathbf r-\mathbf A}{|\mathbf r-\mathbf A|}+\frac{\mathbf r-\mathbf B}{|\mathbf r-\mathbf B|}+\frac{\mathbf r-\mathbf C}{|\mathbf r-\mathbf C|}=0 $$
which leads to $$\mathbf r =\frac{yz\mathbf A + zx\mathbf B + xy \mathbf C}{xy+yz+zx} $$ where the scalers are $x=|\mathbf r-\mathbf A|$, $y=|\mathbf r-\mathbf B|$ and $z=|\mathbf r-\mathbf C|$. Because of the angles all being $2\pi/3$ between them, they satisfy, according to the cosine rule, $$ a^2=y^2+z^2+yz\\ b^2=z^2+x^2+zx\\ c^2=x^2+y^2+xy $$ where $a$, $b$ and $c$ are the side lengths of the triangle, opposite of vertices A, B and C, respectively. Specifically, $a=|\mathbf B-\mathbf C|$, $b=|\mathbf C-\mathbf A|$ and $c=|\mathbf A-\mathbf B|$.
The joint equations above can be worked out, albeit in lengthy and tedious steps. Nonetheless, it produces the following solution $$ \mathbf r =\frac{k_a\mathbf A + k_b\mathbf B + k_c \mathbf C}{(8/\sqrt{3})(a^2+b^2+c^2)K+32K^2} $$ with the coefficients $$ k_a=\left(a^2+ 4K/\sqrt{3}\right)^2 -(b^2-c^2)^2\\ k_b=\left(b^2+ 4K/\sqrt{3}\right)^2 -(c^2-a^2)^2\\ k_c=\left(c^2+ 4K/\sqrt{3}\right)^2 -(a^2-b^2)^2 $$ and $K$ representing the area of the triangle, i.e. $$ K=\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}. $$