If \(\alpha, \beta\), where \(\alpha<\beta\),are the roots of the equation \(\lambda {x}^{2}−\left(\lambda +3\right)x…

Since \(\alpha, \beta\) be the roots

of the quadratic equation \(\lambda x^2-(\lambda+3) x+3=0\)

\(\alpha +\beta =\frac{\lambda +3}{\lambda },\alpha \beta =\frac{3}{\lambda }\)

Now, \(\frac{\beta −\alpha }{\alpha \beta }=\frac{1}{3}\)

\(\Rightarrow \beta −\alpha =\frac{\alpha \beta }{3}=\frac{1}{\lambda }\)

on squaring

\({\alpha }^{2}+{\beta }^{2}−2\alpha \beta =\frac{1}{{\lambda }^{2}}...\left(1\right)\)

\({\alpha }^{2}+{\beta }^{2}+2\alpha \beta =\frac{{(\lambda +3)}^{2}}{{\lambda }^{2}}...\left(2\right)\)

eq(2) –eq(1)

\(4\alpha \beta =\frac{{(\lambda +3)}^{2}−1}{{\lambda }^{2}}\)

\(\Rightarrow \frac{12}{\lambda }=\frac{{\lambda }^{2}+6\lambda +8}{{\lambda }^{2}}\)

\(\Rightarrow {\lambda }^{2}−6\lambda +8=0\)

\(\Rightarrow \lambda =2,4\)

Sum of possible values of \(\lambda\) is \(=6\)