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(C) The sides of the triangle are given by the equations:$(x-y)(x+y)(2x+3y-6)=0$This implies the lines are $L_1: x-y=0$,$L_2: x+y=0$,and $L_3: 2x+3y-6

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$0 < \alpha < 2$

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(C) The sides of the triangle are given by the equations:$(x-y)(x+y)(2x+3y-6)=0$This implies the lines are $L_1: x-y=0$,$L_2: x+y=0$,and $L_3: 2x+3y-6=0$.The vertices of the triangle are the intersection points of these lines:Intersection of $L_1$ and $L_2$: $(0,0)$Intersection of $L_1$ and $L_3$: $x-y=0$ and $2x+3y-6=0 \implies 2x+3x-6=0 \implies 5x=6 \implies x=6/5, y=6/5$Intersection of $L_2$ and $L_3$: $x+y=0$ and $2x+3y-6=0 \implies 2(-y)+3y-6=0 \implies y=6, x=-6$The triangle is formed by vertices $(0,0)$,$(6/5, 6/5)$,and $(-6, 6)$.The point $(0, \alpha)$ lies on the $y$-axis $(x=0)$.For the point $(0, \alpha)$ to be in the interior of the triangle,it must lie between the $x$-axis (which is not a side,but the region is bounded by the lines) and the line $2x+3y-6=0$ along the $y$-axis.At $x=0$,the line $2x+3y-6=0$ gives $3y=6$,so $y=2$.The triangle vertices are $(0,0)$,$(6/5, 6/5)$,and $(-6, 6)$. The interior points on the $y$-axis are between $y=0$ and $y=2$.Thus,$0 < \alpha < 2$.

The combined equation of the three sides of a triangle is $(x^2-y^2)(2x+3y-6)=0$. If the point $(0, \alpha)$ lies in the interior of this triangle,then

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