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Center be O,

From Triangle AOB , $sin\alpha/2 = \dfrac{1}{2r}$

Similarly , $sin\beta/2 = \dfrac{1}{4r} , sin\gamma/2 = \dfrac{3}{2r}$

$\alpha = a , \beta = b , \gamma = c$

$2(a + b + c) = 360^{\circ} , a + b + c = 180^{\circ}$

Then,

$cosa + cosb + cosc = 1 + 4(sina/2)(sinb/2)sinc/2$

$1 - 2sin^{2}a/2 + 1 - 2sin^{2}b/2 + 1 - 2sin^{2}c/2 = 1 + 4(sina/2)(sinb/2)sinc/2$

$3 - 2.\dfrac{1}{4r^{2}} - 2.\dfrac{1}{r^{2}} - 2.\dfrac{9}{4r^{2}} = 1 + \dfrac{3}{r^{3}}$

$3 - \dfrac{14}{2r^{2}} = 1 + \dfrac{3}{r^{3}}$

$\dfrac{3}{r^{3}} + \dfrac{7}{r^{2}} = 2$