If the equation $\left({\mathrm{\lambda }}^2-5\mathrm{\lambda }+6\right){\mathrm{x}}^2+\left({\mathrm{\lambda }}^2-3\mathrm{\lambda }+2\right)\mathrm{x}+\left({\mathrm{\lambda }}^2-4\right)=0$ is satisfied by more than two values of x then $\mathrm{\lambda }=$

$\begin{array}{l}GivenEquation({\lambda }^2-5\lambda +6)+({\lambda }^2-3\lambda +2)x+({\lambda }^2-4)=0\\ IftheEquationa{x}^2+bx+c=0hasmorethen2rootsthenitisanidentityinxand\\ a=0,b=0,c=0\\ \therefore {\lambda }^2-5\lambda +6=0\Rightarrow {\lambda }^2-3\lambda -2\lambda +6=0\\ \Rightarrow (\lambda -3)(\lambda -2)=0\\ \Rightarrow \lambda =2,3\\ and{\lambda }^2-3\lambda +2=0\Rightarrow {\lambda }^2-2\lambda -\lambda +2=0\\ \Rightarrow (\lambda -2)(\lambda -1)=0\\ \Rightarrow \lambda =1,2\\ and{\lambda }^2-4=0\Rightarrow (\lambda +2)(\lambda -2)=0\\ \Rightarrow \lambda =-2,2\\ \therefore Thecommonsolutionis\lambda =2\end{array}$