If $\sqrt{3} +1$ is a root of the equation $3{\mathrm{x}}^3 + {\mathrm{ax}}^2 + \mathrm{bx} + 12 = 0$, where a and b are rational numbers, then (a, b) =

$\begin{array}{l}Let1+\sqrt{3},1-\sqrt{3},\alpha arerootsof3x^3+a{x}^2+bx+12=0\\ \therefore (1+\sqrt{3})(1-\sqrt{3})\alpha =\frac{-12}{3}\\ \Rightarrow (1-3)\alpha =-4\Rightarrow -2\alpha =-4\Rightarrow \alpha =2\\ sumofroots=\frac{-a}{3}\\ \Rightarrow 1+\sqrt{3}+1-\sqrt{3}+2=\frac{-a}{3}\\ \Rightarrow 4=\frac{-a}{3}\Rightarrow a=-12\\ and(1+\sqrt{3})(1-\sqrt{3})+(1+\sqrt{3})2+(1-\sqrt{3})2=\frac{b}{3}\\ \Rightarrow -2+2+2\sqrt{3}+2-2\sqrt{3}=\frac{b}{3}\\ \Rightarrow 2=\frac{b}{3}\Rightarrow b=6\end{array}$