A circle of radius $r$ is inscribed into a triangle. A circle of radius $r$ is inscribed into a triangle $ABC$. Tangent lines to this circle parallel to the sides of the triangle cut out three smaller triangles, $\triangle A_cB_cC$, $\triangle A_bBC_b$, $\triangle AB_aC_a$. The radii of the circles inscribed in these smaller triangles are equal to $1$, $2$ and $3$, respectively. Find $r$.
I have no idea how to start...



Let $|CE|=H_c$ and $|CD|=h_c$ be the altitudes
of similar triangles $\triangle ABC$ and $\triangle A_cB_cC$.
Then
\begin{align}
\frac{|CD|}{r_c}
&=
\frac{|CE|}{r}
\tag{1}\label{1}
,\\
\frac{H_c-2r}{r_c}
&=
\frac{H_c}{r}
\tag{2}\label{2}
,\\
H_c &= \frac{2r^2}{r-r_c}
\tag{3}\label{3}
.
\end{align}
Similarly, two other altitudes of  $\triangle ABC$
in terms of $r,r_a,r_b$ are
\begin{align}
H_a &= \frac{2r^2}{r-r_a}
\tag{4}\label{4}
,\\
H_b &= \frac{2r^2}{r-r_b}
\tag{5}\label{5}
,
\end{align}
and we can apply a well-known relation
\begin{align}
\frac1r&=
\frac1{H_a}+\frac1{H_b}+\frac1{H_c}
\tag{6}\label{6}
\end{align}
to find out that $r$ in terms of $r_a,r_b,r_c$ is just
\begin{align}
r&=r_a+r_b+r_c
\tag{7}\label{7}
.
\end{align}
The original question would be solved by now,
but we can do more than that: we can completely solve the $\triangle ABC$.
Using known  Heron-like formula for the area, we have
\begin{align}
S&=
\frac1{\sqrt{
{(\tfrac1{H_a}+\tfrac1{H_b}+\tfrac1{H_c})}
{(-\tfrac1{H_a}+\tfrac1{H_b}+\tfrac1{H_c})}
{(\tfrac1{H_a}-\tfrac1{H_b}+\tfrac1{H_c})}
{(\tfrac1{H_a}+\tfrac1{H_b}-\tfrac1{H_c})}
}}
\\
&=\frac{r^{7/2}}{\sqrt{r_a r_b r_c}}
\tag{8}\label{8}
.
\end{align}
Next, we can find the semiperimeter $\rho$
and circumradius $R$ of $\triangle ABC$:
\begin{align}
\rho&=\frac Sr
=\frac{r^{5/2}}{\sqrt{r_a r_b r_c}}
\tag{9}\label{9}
,\\
R&=
\frac{2\,S^2}{H_a H_b H_c}
=\tfrac14\,\frac{r(r-r_a)(r-r_b)(r-r_c)}{r_a r_b r_c}
\tag{10}\label{10}
.
\end{align}
Now we are ready to find the three side lengths of $\triangle ABC$
as the roots of cubic equation in terms of $\rho,r,R$:
\begin{align}
x^3-2\rho\,x^2+(\rho^2+r^2+4\,r\,R)\,x-4\,\rho\,r\,R&=0
\tag{11}\label{11}
.
\end{align}
In particular, for $r_a=1,\ r_b=2,\ r_c=3$ we have
\begin{align}
r&=6
,\quad
S=216
,\quad
\rho=36
,\quad
R=15
\tag{12}\label{12}
,
\end{align}
\eqref{11} becomes
\begin{align}
x^3-72\,x^2+1692\,x-12960&=0
\tag{13}\label{13}
\end{align}
with three roots $\{18,\, 24,\, 30\}$,
that is, the sought triangle is
the famous $3-4-5$ right-angled triangle,
scaled by $6$.
Note that the side lengths
are inversely proportional to corresponding radii of incircles.
For another example, the picture illustrates
a solution for $r_a=7,\ r_b=5,\ r_c=3$.
In this case we have $r=15$ and the side lengths are
\begin{align}
a&=\tfrac{120\sqrt7}7
,\quad
b=\tfrac{150\sqrt7}7
,\quad
c=\tfrac{180\sqrt7}7
\tag{14}\label{14}
.
\end{align}

Edit
In fact, solution of the cubic equation \eqref{11}
is unnecessary: since the area and the altitudes are known,
the side lengths can be found explicitly as
\begin{align}
a&=r\,(r-r_a)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}}
\tag{15}\label{15}
,\\
b&=r\,(r-r_b)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}}
\tag{16}\label{16}
,\\
c&=r\,(r-r_c)\,\sqrt{\frac{r}{r_a\,r_b\,r_c}}
\tag{17}\label{17}
.
\end{align}
