Let R be the interior region between the lines 3 x - y + 1 = 0 and x + 2 y - 5 = 0 containing the origin.…

Let, $\mathrm{P}\equiv \left({\mathrm{a}}^2,\mathrm{a}+1\right)$, ${\mathrm{L}}_1:3\mathrm{x}-\mathrm{y}+1=0$ and ${\mathrm{L}}_2:\mathrm{x}+2\mathrm{y}-5=0$.

Now, origin and $\mathrm{P}$ lies on the same side with respect to line ${\mathrm{L}}_1$.

$\Rightarrow {\mathrm{L}}_1\left(0\right)·{\mathrm{L}}_1\left(\mathrm{P}\right)>0$

$\Rightarrow 3\left({\mathrm{a}}^2\right)-(\mathrm{a}+1)+1>0$

$\Rightarrow 3{\mathrm{a}}^2-\mathrm{a}>0$

$\Rightarrow \mathrm{a}\left(3\mathrm{a}-1\right)>0$

$\Rightarrow \mathrm{a}\in (-\infty ,0)\cup \left(\frac{1}{3},\infty \right) ...\left(\mathrm{i}\right)$

Now, ${\mathrm{L}}_2:\mathrm{x}+2\mathrm{y}-5=0$

Origin and $\mathrm{P}$ lies on same side with respect to ${\mathrm{L}}_2$

$\Rightarrow {\mathrm{L}}_2\left(0\right)·{\mathrm{L}}_2\left(\mathrm{P}\right)>0$

$\Rightarrow {\mathrm{a}}^2+2(\mathrm{a}+1)-5<0$

$\Rightarrow {\mathrm{a}}^2+2\mathrm{a}-3<0$

$\Rightarrow (\mathrm{a}+3)(\mathrm{a}-1)<0$

$\therefore \mathrm{a}\in (-3,1) ...\left(\mathrm{ii}\right)$

Using intersection of $\left(i\right) \text{and} \left(ii\right)$,

$\Rightarrow \mathrm{a}\in (-3,0)\cup \left(\frac{1}{3},1\right)$