If $z^3+\left(3+2i\right)z+\left(-1+ia\right)=0$ has one real root, then the value of $a$ lies in the interval $\left(a\in R\right)$

Let $z=\alpha$ be a real root. Then,

${\alpha }^3+\left(3+2i\right)\alpha +\left(-1+i\alpha \right)=0$

$\Rightarrow \left({\alpha }^3+3\alpha -1\right)+i\left(\alpha +2\alpha \right)=0$

$\Rightarrow {\alpha }^3+3\alpha -1=0$ and $\alpha =-a/2$

$\Rightarrow -\frac{a^3}{8}-\frac{3a}{2}-1=0$

$\Rightarrow a^3+12a+8=0$

Let $f\left(a\right)=a^3+12a+8.$

$\therefore f\left(-1\right)<0,f\left(0\right)>0,f\left(-2\right)<0,f\left(1\right)>0,$ and $f\left(3\right)>0$

Hence, $a\in \left(-1,0\right)$ or $a\in \left(-2,1\right)$ or $a\in \left(-2,3\right).$