Find the coordinates of the point in the plane at which the sum of the distances from it to the three points $(0, 0)$, $(2, 0)$, $(0, \sqrt{3})$ is minimal.
To find the coordinates of the point in the plane at which the sum of the distances from it to the three points \((0, 0)\), \((2, 0)\), and \((0, \sqrt{3})\) is minimal, we need to determine the Fermat point of the triangle formed by these vertices. The Fermat point minimizes the total distance to the vertices of the triangle.

1. **Construct Equilateral Triangles:**
   - Construct an equilateral triangle on each of the two sides of the given triangle. Let's choose sides \((0,0)\) to \((2,0)\) and \((0,0)\) to \((0,\sqrt{3})\).

2. **Equilateral Triangle on \((0,0)\) to \((2,0)\):**
   - The third vertex of the equilateral triangle on this side will be at \((1, \sqrt{3})\).

3. **Equilateral Triangle on \((0,0)\) to \((0,\sqrt{3})\):**
   - The third vertex of the equilateral triangle on this side will be at \((-\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2})\).

4. **Draw Lines from New Vertices to Opposite Vertices:**
   - Draw a line from \((1, \sqrt{3})\) to \((0, \sqrt{3})\).
   - Draw a line from \((-\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2})\) to \((2, 0)\).

5. **Find Intersection Point:**
   - The intersection of these two lines is the Fermat point.

Let's calculate the intersection point of these lines.

**Line 1:**
- The line from \((1, \sqrt{3})\) to \((0, \sqrt{3})\) is vertical, so its equation is \(x = 1\).

**Line 2:**
- The line from \((-\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2})\) to \((2, 0)\) has a slope \(m\) given by:
  \[
  m = \frac{0 - \frac{\sqrt{3}}{2}}{2 - (-\frac{\sqrt{3}}{2})} = \frac{-\frac{\sqrt{3}}{2}}{2 + \frac{\sqrt{3}}{2}} = \frac{-\frac{\sqrt{3}}{2}}{\frac{4 + \sqrt{3}}{2}} = \frac{-\sqrt{3}}{4 + \sqrt{3}}
  \]
  Simplify the slope:
  \[
  m = \frac{-\sqrt{3}}{4 + \sqrt{3}} \cdot \frac{4 - \sqrt{3}}{4 - \sqrt{3}} = \frac{-\sqrt{3}(4 - \sqrt{3})}{(4 + \sqrt{3})(4 - \sqrt{3})} = \frac{-4\sqrt{3} + 3}{16 - 3} = \frac{-4\sqrt{3} + 3}{13}
  \]
  The equation of the line is:
  \[
  y - \frac{\sqrt{3}}{2} = \frac{-4\sqrt{3} + 3}{13} \left(x + \frac{\sqrt{3}}{2}\right)
  \]
  Simplify:
  \[
  y = \frac{-4\sqrt{3} + 3}{13} x + \frac{\sqrt{3}}{2} + \frac{-4\sqrt{3} + 3}{13} \cdot \frac{\sqrt{3}}{2}
  \]

6. **Solve for Intersection:**
   - Set \(x = 1\) in the equation of Line 2:
   \[
   y = \frac{-4\sqrt{3} + 3}{13} \cdot 1 + \frac{\sqrt{3}}{2} + \frac{-4\sqrt{3} + 3}{13} \cdot \frac{\sqrt{3}}{2}
   \]
   Simplify:
   \[
   y = \frac{-4\sqrt{3} + 3}{13} + \frac{\sqrt{3}}{2} + \frac{(-4\sqrt{3} + 3)\sqrt{3}}{26}
   \]
   \[
   y = \frac{-4\sqrt{3} + 3}{13} + \frac{\sqrt{3}}{2} + \frac{-12 + 3\sqrt{3}}{26}
   \]
   \[
   y = \frac{-4\sqrt{3} + 3}{13} + \frac{\sqrt{3}}{2} + \frac{-6 + 3\sqrt{3}}{13}
   \]
   \[
   y = \frac{-4\sqrt{3} + 3 - 6 + 3\sqrt{3}}{13} + \frac{\sqrt{3}}{2}
   \]
   \[
   y = \frac{-4\sqrt{3} + 3\sqrt{3} - 3}{13} + \frac{\sqrt{3}}{2}
   \]
   \[
   y = \frac{-\sqrt{3} - 3}{13} + \frac{\sqrt{3}}{2}
   \]
   \[
   y = \frac{-\sqrt{3} - 3 + 6.5\sqrt{3}}{13}
   \]
   \[
   y = \frac{5.5\sqrt{3} - 3}{13}
   \]

Thus, the coordinates of the Fermat point are:
\[
\boxed{\left(1, \frac{5.5\sqrt{3} - 3}{13}\right)}
\]