Figure (Asymptote source)

unitsize(.5cm); defaultpen(linewidth(.8pt)); dotfactor=4;  pair A=(0,0), B=(3,0), C=(0,4);  dot (A); dot (B); dot (C); draw(A--B); draw(A--C); draw(B--C);   draw(Circle(A,1)); draw(Circle(B,2)); draw(Circle(C,3));

The centers of these circles form a 3-4-5 triangle, which has an area equal to 6.
The areas of the three triangles determined by the center and the two points of tangency of each circle are, using Triangle Area by Sine,
$\frac{1}{2} \cdot 1 \cdot 1 \cdot 1 = \frac{1}{2}$
$\frac{1}{2} \cdot 2 \cdot 2 \cdot \frac{4}{5} = \frac{8}{5}$
$\frac{1}{2} \cdot 3 \cdot 3 \cdot \frac{3}{5} = \frac{27}{10}$
which add up to $4.8$. The area we're looking for is the large 3-4-5 triangle minus the three smaller triangles, or $6 - 4.8 = 1.2 = \frac{6}{5} \rightarrow \boxed{(D)}$.