$\sqrt3 +1 \;\text{and}\; -\sqrt3+1 \;\text{are two roots of this equation}$
so $3( \sqrt3 +1 )^3+a(\sqrt3 +1)^2+b(\sqrt3 +1)+12 =0$
$3( 3\sqrt3 +9+3\sqrt3+1 )+a(3+2\sqrt3 +1)+b(\sqrt3 +1)+12 =0$
$a(3+2\sqrt3 +1)+b(\sqrt3 +1)+ 18\sqrt3+42 =0$
$a(4+2\sqrt3)+b(\sqrt3 +1)+ 18\sqrt3+42 =0$ -----(1)
using other root,
$3(-\sqrt3+1 )^3+a(-\sqrt3+1 )^2+b(-\sqrt3+1 )+12=0$
$3(-3\sqrt3+9-3\sqrt3+1 )+a(3-2\sqrt3+1 )+b(-\sqrt3+1 )+12=0$
$a(3-2\sqrt3+1 )+b(-\sqrt3+1 )-18\sqrt3+42=0$
$a(4-2\sqrt3)+b(-\sqrt3+1 )-18\sqrt3+42=0$ -------(2)
using (1) and (2)
$4a+b=-42$ [ by adding ]
$2a+b=-18$ [ by subtracting]
$a=-12\; \text{and}\;b=6$