If $$\alpha, \beta$$, where $$\alpha < \beta$$, are the roots of the quadratic equation  $$\lambda x^{2}-(\lambda + 3)x+3=0$$ and  $$\dfrac{1}{\alpha}-\dfrac{1}{\beta}=\dfrac{1}{3}$$, then the sum of all possible values of $$\lambda$$ is

Given, $$ \lambda x^2 - (\lambda+3)x + 3 = 0 $$ with roots $$ \alpha < \beta$$ , and $$ \dfrac{1}{\alpha} - \dfrac{1}{\beta} = \dfrac{1}{3} $$

Sum of the roots, $$ \alpha + \beta = \dfrac{\lambda+3}{\lambda}, \quad \alpha\beta = \dfrac{3}{\lambda} $$

Now, $$ \dfrac{1}{\alpha} - \dfrac{1}{\beta} = \dfrac{\beta - \alpha}{\alpha\beta} $$

So, $$ \dfrac{\beta - \alpha}{\alpha\beta} = \dfrac{1}{3} $$

$$  \Rightarrow \beta - \alpha = \dfrac{\alpha\beta}{3} = \dfrac{1}{\lambda} $$

Now using, $$ (\beta - \alpha)^2 = (\alpha + \beta)^2 - 4\alpha\beta $$

$$ \left(\dfrac{1}{\lambda}\right)^2 = \left(\dfrac{\lambda+3}{\lambda}\right)^2 - \dfrac{12}{\lambda} $$

or, $$ 1 = (\lambda+3)^2 - 12\lambda $$

or, $$ 1 = \lambda^2 + 6\lambda + 9 - 12\lambda = \lambda^2 - 6\lambda + 9 $$

or, $$ \lambda^2 - 6\lambda + 8 = 0$$

or, $$ (\lambda - 2)(\lambda - 4) = 0$$

or, $$ \lambda = 2, 4 $$

On putting the values and checking the discriminants, we observe that both give real and distinct roots, so both are valid.

So, the sum of all the values of $$\lambda$$ is $$2 + 4 = \boxed{6} $$