Given lines are
$$
\begin{aligned}
& 3 x^2-4 x y+y^2=0 \\
& \Rightarrow 3 x^2-3 x y-x y+y^2=0 \\
&(3 x-y)(x-y)=0 \\
& \Rightarrow 3 x-y=0, x-y=0 \\
& \text { and } 2 x-y=6
\end{aligned}
$$
The point of intersection of these lines are $(0,0),(-6,-18)$ and $(6,6)$.
$$
\begin{aligned}
\therefore \text { Area of triangle } & =\frac{1}{2}\left|\begin{array}{lll}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{array}\right| \\
& =\frac{1}{2}\left|\begin{array}{ccc}
0 & 0 & 1 \\
-6 & -18 & 1 \\
6 & 6 & 1
\end{array}\right| \\
& =\frac{1}{2}(36+108)=\frac{1}{2} \\
& =36 \text { sq unit }
\end{aligned}
$$