If the equation $\left|{\mathrm{x}}^2-5\mathrm{x}+6\right|-\mathrm{\lambda x}+7\mathrm{\lambda }=0$ has exactly three solutions then $\lambda$ is equal to

$\begin{array}{l}\left|{\mathrm{x}}^2-5\mathrm{x}+6\right|-\mathrm{\lambda x}+7\mathrm{\lambda }=0\text{ has there solutions }\\ \left|{\mathrm{x}}^2-5\mathrm{x}+6\right|=\mathrm{\lambda x}-7\mathrm{\lambda }\\ \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-5\mathrm{x}+6\\ -\left({\mathrm{x}}^2-5\mathrm{x}+6\right)=\mathrm{\lambda x}-7\mathrm{\lambda }=0\\ {\mathrm{b}}^2-4\mathrm{ac}=0\\ (\mathrm{\lambda }-5{)}^2-4(6-7\mathrm{\lambda })=0\end{array}$
$\begin{array}{l}\Rightarrow {\mathrm{\lambda }}^2-10\mathrm{\lambda }+25-24+28\mathrm{\lambda }=0\\ \Rightarrow {\mathrm{\lambda }}^2+18\mathrm{\lambda }+1=0\\ \mathrm{\lambda }=\frac{-15\pm \sqrt{\left({18}^2\right)-4}}{2}=-9\pm \sqrt{80}=-9\pm 4\sqrt{5}\end{array}$