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

Rewrite the given condition: $\frac{1}{\alpha }-\frac{1}{\beta }=\frac{\beta -\alpha }{\alpha \beta }=\frac{1}{3}$, so $\beta -\alpha =\frac{\alpha \beta }{3}=\frac{1}{\lambda }$.

Square: $(\beta -\alpha {)}^2=(\alpha +\beta {)}^2-4\alpha \beta$:

$\frac{1}{{\lambda }^2}=\frac{(\lambda +3{)}^2}{{\lambda }^2}-\frac{12}{\lambda }$

Multiply through by ${\lambda }^2$: $1=(\lambda +3{)}^2-12\lambda \Rightarrow 1={\lambda }^2+6\lambda +9-12\lambda$

$\Rightarrow {\lambda }^2-6\lambda +8=0\Rightarrow (\lambda -2)(\lambda -4)=0$.

So $\lambda \in \{2,4\}$. Note $\lambda =0$ would make the equation linear (and the working assumed $\lambda \ne 0$).

Sum of valid values: $2+4=6$. Option (1).