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(D) The given equation of the pair of straight lines is $x^{2}-3xy+\lambda y^{2}+3x-5y+2=0$.Comparing this with the general form $ax^{2}+2hxy+by^{2}+2

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(D) The given equation of the pair of straight lines is $x^{2}-3xy+\lambda y^{2}+3x-5y+2=0$.Comparing this with the general form $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$,we get $a=1$,$2h=-3 \Rightarrow h=-\frac{3}{2}$,and $b=\lambda$.The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{h^{2}-ab}}{a+b} ight|$.Given $	an 	heta = \frac{1}{3}$,we have $\frac{1}{3} = \left| \frac{2\sqrt{\frac{9}{4}-\lambda}}{1+\lambda} ight|$.Squaring both sides: $\frac{1}{9} = \frac{4(\frac{9}{4}-\lambda)}{(1+\lambda)^{2}} = \frac{9-4\lambda}{(1+\lambda)^{2}}$.$(1+\lambda)^{2} = 9(9-4\lambda) = 81-36\lambda$.$1+2\lambda+\lambda^{2} = 81-36\lambda$.$\lambda^{2}+38\lambda-80=0$.$(\lambda+40)(\lambda-2)=0$.Since $\lambda \geq 0$,we have $\lambda=2$.

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

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