Let the centroid of an equilateral triangle $$ABC$$ be at the origin. Let one of the sides of the equilateral triangle be along the straight line $$x + y = 3$$. If $$R$$ and $$r$$ be the radius of circumcircle and incircle respectively of $$\triangle ABC$$, then $$(R + r)$$ is equal to :

The centroid of equilateral triangle $$ABC$$ is at the origin and one side lies along $$x + y = 3$$. The distance from the origin to the line $$x + y = 3$$ is $$\frac{|0 + 0 - 3|}{\sqrt{1^2 + 1^2}} = \frac{3}{\sqrt{2}}$$.

For an equilateral triangle, the centroid coincides with both the circumcenter and incenter. The distance from the centroid to a side equals the inradius $$r$$, so $$r = \frac{3}{\sqrt{2}}$$. For an equilateral triangle, the circumradius is $$R = 2r$$, so $$R = \frac{6}{\sqrt{2}}$$.

Therefore $$R + r = \frac{6}{\sqrt{2}} + \frac{3}{\sqrt{2}} = \frac{9}{\sqrt{2}}$$.