The answer is 575157.

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Label by ${{r}_{1}}$ , ${{r}_{2}}$ and ${{r}_{3}}$ the radii of the incircles of $\triangle AED$ , $\triangle EDB$ and $\triangle BCE$ respectively. Let $x=AD$ , $y=AE$ , $z=EB$ , $w=ED$ . Then $DB=4-x$ and $EC=5-y$ . For the radii we have

${{r}_{1}}=\dfrac{\text{Area of }\triangle \text{AED}}{\text{Semiperimeter of }\triangle \text{AED}}=\dfrac{\dfrac{1}{2}AD\cdot AE\cdot \sin A}{\dfrac{1}{2}\left( x+y+w \right)}=\dfrac{\dfrac{1}{2}xy\cdot \dfrac{3}{5}}{\dfrac{1}{2}\left( x+y+w \right)} \ \ \ \ \ (1)$

Similarly,

${{r}_{2}}=\dfrac{\dfrac{1}{2}\left( 4-x \right)y\cdot \dfrac{3}{5}}{\dfrac{1}{2}\left( 4-x+w+z \right)} \ \ \ \ \ and \ \ \ \ \ {{r}_{3}}=\dfrac{\frac{1}{2}\cdot 3\left( 5-y \right)\cdot \dfrac{4}{5}}{\dfrac{1}{2}\left( 5-y+3+z \right)}$

By cosine rule on $\triangle AED$ ,

$E{{D}^{2}}=A{{D}^{2}}+A{{E}^{2}}-2AD\cdot AE\cdot \cos A\Rightarrow {{w}^{2}}={{x}^{2}}+{{y}^{2}}-2xy\cdot \dfrac{4}{5}$ By cosine rule on $\triangle AEB$ ,

$E{{B}^{2}}=A{{E}^{2}}+A{{B}^{2}}-2AE\cdot AB\cdot \cos A\Rightarrow {{z}^{2}}={{y}^{2}}+16-2y\cdot 4\cdot \dfrac{4}{5}$ Since ${{r}_{1}}={{r}_{2}}$ and ${{r}_{1}}={{r}_{3}}$ , we get a system of four equations in four variables:

$\begin{aligned} \dfrac{x}{x+y+w} & =\dfrac{4-x}{4-x+w+z} \\[0.5em] \dfrac{xy}{x+y+w} & =\dfrac{4\left( 5-y \right)}{8-y+z} \\[0.5em] {{w}^{2}} & ={{x}^{2}}+{{y}^{2}}-\dfrac{8}{5}xy \\[0.5em] {{z}^{2}} & ={{y}^{2}}+16-\dfrac{32}{5}y \\ \end{aligned}$

The solution of the system is $x \approx 2.18350128379179703 \ \ \ \ \ y \approx 3.29739514976766282 \ \ \ \ \ z \approx 2.40197539854143083 \ \ \ \ \ w \approx 2.029952256023100340$ Substituting in $(1)$ we can find the (common) radius $R$ of the three incircles: $R \approx 0.5751573629$ Hence, the answer is $\left\lfloor {{10}^{6}}R \right\rfloor =\boxed{575157}$ .