Given equation of pair of lines is
$$
\begin{aligned}
& x^2-3 x y+\lambda y^2+3 x-5 y+2=0 \\
& \text { Here, } \mathrm{a}=1, \mathrm{~b}=\lambda, \mathrm{c}=2, \mathrm{f}=\frac{-5}{2}, \mathrm{~g}=\frac{3}{2}, \mathrm{~h}=\frac{-3}{2} \\
& \theta=\tan ^{-1}\left(\frac{1}{3}\right) \Rightarrow \tan \theta=\frac{1}{3}
\end{aligned}
$$
Since $\tan \theta=\left|\frac{2 \sqrt{\mathrm{h}^2-a b}}{\mathrm{a}+\mathrm{b}}\right|$
$$
\begin{aligned}
& \Rightarrow \frac{1}{3}=\left|\frac{2 \sqrt{\left(\frac{-3}{2}\right)^2-\lambda}}{\lambda+1}\right| \\
& \Rightarrow(\lambda+1)^2=9(9-4 \lambda) \Rightarrow \lambda^2+38 \lambda-80=0 \\
& \Rightarrow(\lambda+40)(\lambda-2)=0 \Rightarrow \lambda=2 \quad \ldots[\because \lambda \geq 0]
\end{aligned}
$$