Answer
True.
Work Step by Step
The general equation of a conic in the form of $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$
(i) defines a parabola if $B^2-4AC=0$
(ii) defines an ellipse (or a circle) if $B^2-4AC\lt0$
(iii) defines a hyperbola if $B^2-4AC\gt0$
Hence here the answer since $A=3, B=-12, C=12$, hence $B^2-4AC=(-12)^2-4=(3)(12)=144-144=0$. Thus it defines a parabola.
Hence the statement is true.
