The range of $\alpha$, if  $(\alpha , {\alpha }^2)$ lies inside the triangle having sides along the lines $2x+3y=1, x+2y-3=0, 6y=5x-1$

Let A, B, C be vertices of the triangle. 
$A=(-7,5), B=\left(\frac{{\displaystyle 5}}{{\displaystyle 4}},\frac{{\displaystyle 7}}{{\displaystyle 8}}\right)$
$C=\left(\frac{{\displaystyle 1}}{{\displaystyle 3}},\frac{{\displaystyle 1}}{{\displaystyle 9}}\right)$. Sign of A w.r.t. BC to -ve.
If P lies inside the triangle ABC, then sign of P will be the same as sign of A w.r.t. the line BC
$\Rightarrow 5\alpha -6{\alpha }^2-1<0$.....(i)
similarly $2\alpha +3{\alpha }^2-1>0$.....(ii)
And $\alpha +2{\alpha }^2-3<0$.....(iii)
Solving (i), (ii) and (iii) for  and then taking intersection,
we get $\alpha \in \left(\frac{{\displaystyle 1}}{{\displaystyle 2}},1\right)\cup \left(-\frac{{\displaystyle 3}}{{\displaystyle 2}},-1\right)$