$\text{ Let }\lambda \ne 0\text{ be in }\mathrm{R}\text{ . If }\alpha \text{ and }\beta \text{ are the roots of the equation, }{\mathrm{x}}^2-\mathrm{x}+2\lambda =0\text{ and }\alpha \text{ and }\gamma \text{ are the }$

$\text{ roots of the equation, }3{\mathrm{x}}^2-10\mathrm{x}+27\lambda =0\text{ , then }\frac{\beta \gamma }{\lambda }\text{ is equal to: }$

$\begin{array}{l}3x^2-3x+6\lambda =0\\ 3x^2-10x+27\lambda =0\\ - + -\\ 7x-21\lambda =0\\ x=3\lambda =\alpha \\ \alpha \beta =2\lambda \Rightarrow \beta =\frac{2}{3}\\ \alpha \gamma =9\lambda \Rightarrow \gamma =3\end{array}$

$\begin{array}{ll}\text{ Now }\beta \gamma =2& \text{ also }\alpha =3\lambda \text{ satisfies }{x}^2-x+2\lambda =0\\ \Rightarrow 9{\lambda }^2-3\lambda +2\lambda =0& 9{\lambda }^2-\lambda =0\\ \lambda =\frac{1}{9}, & \text{ So }\frac{\beta \gamma }{\lambda }=\frac{2}{1/9}=18\end{array}$