A circle and equilateral triangle inscribed in a semicircle.

$m\angle{CFD} = 60^{\circ} \implies m\angle{AFC} = 120^{\circ}$ and letting $m\angle{NOE} = \theta \implies$

$120^{\circ} = \dfrac{1}{2}(m(\widehat{NBE}) - m(\widehat{NRE})) = \dfrac{1}{2}(360^{\circ} - 2\theta) \implies$

$240^{\circ} = 360^{\circ} - 2\theta \implies \theta = 60^{\circ} \implies m\angle{NOF} =30^{\circ} \implies$

$\angle{OFN} = 60^{\circ} \implies \dfrac{r}{y} = \tan(60^{\circ}) = \sqrt{3}$ , where $y = \overline{NF}$

$\implies y = \dfrac{r}{\sqrt{3}} \implies m = \dfrac{2r}{\sqrt{3}} \implies \overline{BF} = r + m = \dfrac{2 + \sqrt{3}}{\sqrt{3}}r = 1 \implies$

$r = \dfrac{\sqrt{3}}{2 + \sqrt{3}} = 2\sqrt{3} - 3 \implies A_{c} = \pi(2\sqrt{3} - 3)^2 = 3(7 - 4\sqrt{3})\pi$

$\alpha(\beta - \lambda\sqrt{\alpha})\pi \implies \alpha + \beta + \lambda = \boxed{14}$

Let the equilateral triangle be $ABC$ , where $C$ is the center of the semicircle, the tangent points to equilateral triangle, the semicircle diameter, and the semicircle arc be $M$ , $N$ , and $Q$ respectively, and the radius of the circle be $r$ . Then $OM=ON=OQ=r$ , where $O$ is the center of the circle, and they are perpendicular to the respective line.

We note that $C$ , $O$ , and $Q$ are colinear. Since $\triangle ABC$ is equilateral, $\angle ACB = 60^\circ$ , $\implies \angle ACN = 180^\circ - 60^\circ = 120^\circ$ . Due to symmetry, $\angle OCM = \angle OCN = 60^\circ$ . Then

$\begin{aligned} QO+OC& = QC \\ QO + ON \csc \angle OCN & = QC \\ r + r \csc 60^\circ & = 1 \\ r & = \frac 1{1+\csc 60^\circ} = \frac 1{1+\frac 2{\sqrt 3}} = \frac {\sqrt 3}{\sqrt 3 + 2} = \frac {\sqrt 3 (2-\sqrt 3)}{4-3} = 2\sqrt 3 - 3 \\ \implies A_c & = \pi r^2 = (2\sqrt 3 - 3)^2 \pi = (21 - 12\sqrt 3)\pi = 3 (7 - 4 \sqrt 3)\pi \end{aligned}$

Therefore $\alpha + \beta + \gamma = 3+7+4 = \boxed{14}$ .