If the angle between the lines given by the equation
\(x^{2}-3 x y+\lambda y^{2}+3 x-5 y+2=0, \lambda \geq 0\), is \(\tan ^{-1}\left(\frac{1}{3}\right)\), then \(\lambda=\)

(D)
Given equation of pair of straight lines be \(x^{2}-3 x y+\lambda y^{2}+3 x-5 y+2=0\).
Here \(\mathrm{a}=1, \mathrm{~b}=\lambda, \mathrm{h}=\frac{-3}{2}\) and angle \(\theta\) is given by \(\tan ^{-1}\left(\frac{2 \sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right)\)
\(\therefore \frac{1}{3}=\frac{2 \sqrt{\frac{9}{4}-\lambda}}{1+\lambda} \Rightarrow(1+\lambda)^{2}=\left(6 \sqrt{\frac{9}{4}-\lambda}\right)^{2}\)
\(\therefore 1+2 \lambda+\lambda^{2}=36\left(\frac{9}{4}-\lambda\right)=81-36 \lambda\)
\(\therefore \lambda^{2}+38 \lambda-80=0 \Rightarrow(\lambda+40)(\lambda-2)=0\)
\(\therefore \lambda=2,-40\)