The equation $\left({\mathrm{a}}^2-\mathrm{a}-2\right){\mathrm{x}}^2+\left({\mathrm{a}}^2-4\right)\mathrm{x}+\left({\mathrm{a}}^2-3\mathrm{a}+2\right)=0$ will have more than two solutions if a equal to

Given , $\left({\mathrm{a}}^2-\mathrm{a}-2\right){\mathrm{x}}^2+\left({\mathrm{a}}^2-4\right)\mathrm{x}+\left({\mathrm{a}}^2-3\mathrm{a}+2\right)=0-----\left(\mathrm{i}\right)$

Since, Eq.(i) have more than two solutions.
So, it is an Identity.

$\begin{array}{l}\therefore {\mathrm{a}}^2-\mathrm{a}-2={\mathrm{a}}^2-4={\mathrm{a}}^2-3\mathrm{a}+2=0\\ \left[\because \text{ if }{\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0\text{ is an identity, then }\mathrm{a}=\mathrm{b}=\mathrm{c}=0\right]\end{array}$

$\begin{array}{l}{\mathrm{a}}^2-\mathrm{a}-2=0\Rightarrow a=2,-1\\ and a^2-4=0\Rightarrow a=\pm 2,\\ and a^2-3a+2=0\Rightarrow a=1,2\\ therefore common value of a=2\end{array}$