Three circles having centres on the three sides of a triangle
Here's a (not-to-scale) picture of the situation:
Necessarily, each circle center ($D$, $E$, or $F$) is the point where an angle bisector meets an opposite edge; moreover, the points of tangency of a circle with the adjacent edges (for instance, $D^\prime$ and $D^{\prime\prime}$) are simply the feet of perpendiculars from the center to those edges.
We'll write $a$, $b$, $c$ for the lengths of sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$, and $d$, $e$, $f$ for the radii of $\bigcirc D$, $\bigcirc E$, $\bigcirc F$. Now, observing that each angle bisector cuts the triangle with into sub-triangles with convenient "bases" and "heights", we can compute the area, $T$, of $\triangle ABC$ in three ways:
$$T \;=\; \frac12 d\;(b+c) \;=\; \frac12 e\;(c+a) \;=\; \frac12 f\;(a+b) \tag{1}$$
Of course, writing $r$ for the inradius of $\triangle ABC$, we have a well-known fourth formula for area: $$T \;=\; \frac12 r\;(a+b+c) \tag{2}$$
We can easily eliminate $a$, $b$, $c$ from the above. For instance, $$\begin{align} b+c = \frac{2T}{d}\quad c+a=\frac{2T}{e}\quad a+b = \frac{2T}{f} &\quad\to\quad 2(a+b+c) = 2T\left(\;\frac{1}{d}+\frac{1}{e}+\frac{1}{f}\;\right) \tag{3} \\ a+b+c = \frac{2T}{r} &\quad\to\quad 2(a+b+c) = 2T\left(\;\frac{2}{r}\;\right) \tag{4} \end{align}$$ so that, as @Jack notes,
$$\frac{2}{r} = \frac{1}{d} + \frac{1}{e} + \frac{1}{f} \tag{$\star$}$$
Addendum (four years later!). As @jmerry has observed, the specific configuration in the problem statement is invalid. If we solve $(1)$ for $a$, $b$, $c$, we find $$a = \left(-\frac1d + \frac1e + \frac1f \right)T \qquad b = \left(\frac1d - \frac1e + \frac1f \right)T \qquad c = \left(\frac1d + \frac1e - \frac1f \right)T$$ With $d=18$, $e=6$, $f=9$, these become $a=2T/9$, $b=0$, $c=T/9$, which do not correspond to a valid triangle ... not even a validly-degenerate one. (It's a good thing I didn't claim my picture was to scale.) A valid triangle requires that the three aspects of the Triangle Inequality hold $$a \leq b+c \qquad b \leq c+a \qquad c \leq a+b$$ which, in turn, require $$\frac3d \geq \frac1e + \frac1f \qquad \frac3e \geq \frac1f+\frac1d \qquad \frac3f \geq \frac1d+\frac1e$$ (The first of these is violated by the given values of $d$, $e$, $f$.)
The existing answers have provided the intended solution. On the other hand, there's a catch - the problem has a fatal flaw.
What can we say about the side lengths of the triangle? We have $b+c$, $a+c$, and $a+b$ in proportions $1:3:2$, so $a$, $b$, and $c$ are in proportions $2:0:1$. That is not a triangle; $a>b+c$.
Let the sides be $a,b,c$, the radii of the circles opposite them be $R_A,R_B,R_C$, and the area of the triangle be $T$, mixing notation from the answers of Blue and Jack D'Aurizio. The triangle inequality $a<b+c$ gives $$\frac1{R_B}+\frac1{R_C}=\frac1{2T}(2a+b+c) < \frac1{2T}(3b+3c) = \frac3{R_A}$$ $$R_A < \frac{3R_BR_C}{R_B+R_C}$$ and similarly for the other two orders; the largest radius must be less than $\frac32$ times the harmonic mean of the other two. An acute triangle would be an even stronger restriction. For the given trio of radii $18,6,9$, this is not true.
As such, the true answer to the problem:
The given configuration is impossible. There is no such triangle.