Triangle ABC is formed by the lines $x-y+2=0, 3x+2y-2=0$ and $2x+y+1=0$. Then, 

Lines are

$\begin{array}{r}x-y+2=0\\ 3x+2y-2=0\\ 2x+y+1=0\end{array}\begin{array}{r} -\left(1\right)\Rightarrow \text{ side }BC\\ -\left(2\right)\Rightarrow \text{ side }CA\\ -\left(3\right)\Rightarrow \text{ side }AB\end{array}$

Solving the equations we get $A$ is (-4, 7) B as (-1, 1)

and $C$ as $\left(\frac{-2}{5},\frac{8}{5}\right)$

If $P (a, 2)$ is to lie inside the triangle $ABC,$ both $A$ and $P$ must lie on the same side of $BC,$ both $B$ and $P$ must lie on the same side of $AB$

$\begin{array}{l}x-y+2<0\text{ for }x=-4,y=7\\ \Rightarrow \alpha -2+2<0\\ \Rightarrow \alpha <0\end{array}$

Again, $3x+2y-2<0\text{ for }x=-1,y=1$

Lastly, $2x+y+1>0\text{ for }x=\frac{-2}{5},y=\frac{8}{5}$

From (4), (5) and (6) we get 

$\frac{-3}{2}<\alpha <\frac{-2}{3}$

Note that all the vertices are in the second quadrant. Hence, if $a>0,$ the point $(a, 2)$ lies in the first quadrant