The equation of a circle is given as \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Expanding this, we have \(x^{2} + y^{2} - 2ax - 2by + a^{2} + b^{2} = r^{2}\)

Comparing with the given equation, \(3x^{2} + 3y^{2} + 6x -12y + 6 = 0 \equiv x^{2} + y^{2} + 2x - 4y + 2 = 0\) (making the coefficients of \(x^{2}\) and \(y^{2}\) = 1 , we get that

\(-2a = 2 \implies a = -1\)

\(2b = 4 \implies b = 2\)

\(r^{2} - a^{2} - b^{2} = -2\)

\(\therefore r^{2} - (-1)^{2} - (2)^{2} = -2 \implies r^{2} = -2 + 1 + 4 = 3\)

\(r = \sqrt{3}\)