Question

$\frac{\beta - \alpha}{\alpha\beta} = \frac{1}{3},\ \alpha + \beta = \frac{\lambda + 3}{\lambda},\alpha\beta = \frac{3}{\lambda}$

$$\beta - \alpha = \frac{\alpha\beta}{3} = \frac{1}{\lambda} $$on squaring

$$\begin{array}{r} \begin{matrix} \alpha^{2} + \beta^{2} - 2\alpha\beta = \frac{1}{\lambda^{2}} & \ldots\ldots\ldots(1) \\ \alpha^{2} + \beta^{2} + 2\alpha\beta = \frac{(\lambda + 3)^{2}}{\lambda^{2}} & \ldots\ldots\ldots\ldots(2) \\ (2) - (1) & 4\alpha\beta = \frac{(\lambda + 3)^{2} - 1}{\lambda^{2}} \\ & \frac{12}{\lambda} = \frac{\lambda^{2} + 6\lambda + 8}{\lambda^{2}} \\ \Rightarrow \lambda^{2} - 6\lambda^{2} + 8\lambda = 0 & \\ \Rightarrow \lambda = 0,2,4 & \end{matrix}\#(2) \end{array}$$

Sum of possible values of $\lambda$ is $= 6$