Line 1: $2x-3y+6=0$. When $x=0, y=2$; when $y=0, x=-3$. Points: (0,2), (-3,0). Line 2: $2x+3y-18=0$. When $x=0, y=6$; when $y=0, x=9$. Points: (0,6), (9,0). Line 3: $x=0$ is the y-axis. Intersection of Line 1 and Line 2: Adding both equations: $(2x-3y+6)+(2x+3y-18)=0 \Rightarrow 4x-12=0 \Rightarrow x=3$. Substitute in Line 1: $2(3)-3y+6=0 \Rightarrow 12-3y=0 \Rightarrow y=4$. Point: (3,4). Intersection of Line 1 and x=0: (0,2). Intersection of Line 2 and x=0: (0,6). So the vertices of the triangle are: $(0,2)$, $(0,6)$, and $(3,4)$. Area $=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)| =\frac{1}{2}|0(6-4)+0(4-2)+3(2-6)| =\frac{1}{2}|0+0-12| = \frac{1}{2}\times 12 = 6$ sq. units.