I like big circles

The key to answering this question correctly is to first of all recognize that the distance of the point of concurrency of tangents P is equidistant from all the 3 points of contact of the circles D , E and F . Since this point is symmetrically located with respect to the points D , E and F , it has to be the incentre of the $\Delta ABC$ .

Therefore the required distance = In radius of the $\Delta ABC$ .

Let S be the semi-perimeter of $\Delta ABC$ ; $2 \times S = 21 + 32 + 32 + 19 + 19 + 21$ $\Rightarrow S = 72$

Area of $\Delta ABC : \Delta$ = $\sqrt{72 \times (72-53) \times (72-51) \times (72-40)}$

Inradius = $\frac{ \Delta}{S}$ (and after putting the values) =13.31665 =13.317 .

Let P be the meeting point of the tangents.
The three centers of the circles will form a centers triangle.
P, corresponding two points of tangency, and the vertex form a kite.
So the line joining P to the vertex is an angle bisector.
Thus P is the incenter of the triangle.
We therefore desire to find the distance of P from the points of tangency.
This is the inradius r equidistance from all points of tangency .
Centers triangle has sides 19+20=40, 21+32=53, 32+19=51.
$\therefore\ r=\dfrac{area} s.\ \ \ \ \ \ \ \ s=19+21+32=72. \\ r=\dfrac {\sqrt {72*32*19*21}}{72}=13.3166.\\ P\ is\ at\ \ \color{#D61F06}{13.317}\ \ \text{from each point of tangency.}$