Squares and Circles!!

For red circle:

For $\overline{AC}: y = x$ and $m_{\overline{ED}} = -2 \implies y = -2x + 2a \implies x = y = \dfrac{2a}{3}$

$\implies G(\dfrac{2a}{3},\dfrac{2a}{3}) \implies \overline{AG} = \dfrac{2\sqrt{2}}{3}a$ and $\overline{GD} = \dfrac{\sqrt{5}}{3}a$

$\implies A_{\triangle{AGD}} = \dfrac{1}{2}(a)(\dfrac{2a}{3}) = \dfrac{a^2}{3} =$ $\dfrac{ar}{2}(\dfrac{3 + 2\sqrt{2} + \sqrt{5}}{3}) \implies$

$2a^2 - (3 + 2\sqrt{2} + \sqrt{5})ar = 0 \implies a(2a - (3 + 2\sqrt{2} + \sqrt{5})r) = 0$ and

$a \neq 0 \implies\boxed{r = \dfrac{2a}{\sqrt{5} + 2\sqrt{2} + 3}}$

For green circle:

$h_{\triangle{EGC}} = a - \dfrac{2a}{2} = \dfrac{a}{3}, \overline{EG} = \dfrac{\sqrt{5}}{6}a$ and $\overline{GC} = \dfrac{\sqrt{2}}{3}a \implies$

$A_{\triangle{EGC}} = \dfrac{1}{2}(\dfrac{a}{2})(\dfrac{a}{3}) = \dfrac{a^2}{12} =$ $\dfrac{1}{2}(a r_{2})(\dfrac{\sqrt{5} + 2\sqrt{2} + 3}{6}) \implies$

$a(a - (\sqrt{5} + 2\sqrt{2} + 3)r_{2}) = 0$ and $a \neq 0 \implies \boxed{r_{2} = \dfrac{a}{\sqrt{5} + 2\sqrt{2} + 3}}$

Similarly using the same method for the pink circle we have:

$\boxed{r_{3} = \dfrac{a}{\sqrt{5} + \sqrt{2} + 3}}$

$\implies \dfrac{r + r_{2} + r_{3}}{a} =\dfrac{12 + 5\sqrt{2} +4\sqrt{5}}{(3 + 2\sqrt{2} + \sqrt{5})(3 + \sqrt{2} + \sqrt{5})} =$

$\dfrac{3 * 2^2 + 5\sqrt{2} + 2^2\sqrt{5}}{(3 + 2\sqrt{2} + \sqrt{5})(3 + \sqrt{2} + \sqrt{5})} =$ $\dfrac{\alpha * \beta^{\beta} + \lambda\sqrt{\beta} + \beta^{\beta}\sqrt{\lambda}}{(\alpha + \beta\sqrt{\beta} + \sqrt{\lambda})(\alpha + \sqrt{\beta} + \sqrt{\lambda})}$

$\implies \alpha + \beta + \lambda = \boxed{10}$ .