The value of a for which $\frac{{\mathrm{x}}^3-6{\mathrm{x}}^2+11\mathrm{x}-6}{{\mathrm{x}}^3+{\mathrm{x}}^2-10\mathrm{x}+8}+\frac{\mathrm{\alpha }}{30}=0$ does not have a real solution is/are

$f\left(x\right)=\frac{x^3-6x^2+11x-6}{x^3+x^2-10x+8}=\frac{(x-1)(x-2)(x-3)}{(x-1)(x-2)(x+4)}$
$\therefore f\left(x\right)\text{ is not defined for }x=1,2,-4\text{. }$
$\therefore \text{ For }x\ne 1,2,-4,f\left(x\right)=\frac{x-3}{x+4}=1-\frac{7}{x+4}$
$\begin{array}{l}\therefore f\left(x\right)\ne -\frac{2}{5},-\frac{1}{6},1\\ \therefore f\left(x\right)+\frac{\alpha }{30}=0\text{ does not have a real solution if }\\ \frac{\alpha }{30}=-1,+\frac{2}{5},+\frac{1}{6}\end{array}$
$\text{ i.e., }\alpha =-30,12,5$