How to find a point on a line that minimizes sum of distances from three given points?

As JJacquelin answered, Newton method is the simplest way to solve your problem.

As usual, you need a guess. You can get it minimizing first the sum of the squared distances that is to say $$\frac{d}{dx} \sum_{i=1}^3 \big({(x-x_i)^2+(0 - y_i)^2}\big) = 0\implies x=\frac 13\sum_{i=1}^3 x_i$$ For the test case, this gives $x_0=6.83$.

Now, for the real problem, Newton method will generate the following iterates $$\left( \begin{array}{cc} n & x_{(n)} \\ 0 & 6.8333333 \\ 1 & 6.3992290 \\ 2 & 6.4017089 \\ 3 & 6.4017090 \end{array} \right)$$

Edit

Making the problem more general for $n$ points, the iterates of Newton method are given by

$$x_{(n+1)}=x_{(n)}-\frac{\sum _{i=1}^n \frac{x_{(n)}-x_i}{\left(x_{(n)}^2-2 x_{(n)} x_i+x_i^2+y_i^2\right)^{1/2}}}{\sum _{i=1}^n \frac{y_i^2}{\left(x_{(n)}^2-2 x_{(n)} x_i+x_i^2+y_i^2\right)^{3/2}}}\qquad \text{with} \qquad x_{(0)}=\frac 1n\sum_{i=1}^n x_i$$ which should converge in a couple of iterations.