If the lines $x+y=0,x-y=0$  and $2x+3y-6=0$ are the sides of triangle and $(-2,a)$  is an interior point of the triangle then a lies 

$\begin{array}{l}(x+y)(x-y)(2x+3y-6)=0\\ x+y=0\cdots \left(1\right),x-y=0\cdots \left(2\right)\\ 2x+3y-6=0\cdots \left(3\right)\end{array}$

From equations $\left(1\right)\left(2\right)\left(3\right)$

$A(\frac{6}{5},\frac{6}{5}),B(0,0),C(-6,6)$

From the line $\overline{BC}\Rightarrow \frac{6}{5}+\frac{6}{5}=\frac{12}{5}>0$

$\begin{array}{l}\Rightarrow -2+a>0\\ a>2\cdots \left(4\right)\end{array}$

From the line $\overline{AC}\Rightarrow 2\left(0\right)+3\left(0\right)-6=-6<0$

$\begin{array}{l}\Rightarrow -4+3a-6<0\\ \Rightarrow 3a<10\\ a<\frac{10}{3}\cdots \left(5\right)\end{array}$