Let the circles $$C_1 : (x - \alpha)^2 + (y - \beta)^2 = r_1^2$$ and $$C_2 : (x - 8)^2 + \left(y - \frac{15}{2}\right)^2 = r_2^2$$ touch each other externally at the point $$(6, 6)$$. If the point $$(6, 6)$$ divides the line segment joining the centres of the circles $$C_1$$ and $$C_2$$ internally in the ratio $$2 : 1$$, then $$(\alpha + \beta) + 4(r_1^2 + r_2^2)$$ equals

A

125

B

130

C

110

D

145