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(A) The given equation is ${x^2} - 3xy + \lambda {y^2} + 3x - 5y + 2 = 0$. Comparing this with $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a = 1$

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$2$

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(A) The given equation is ${x^2} - 3xy + \lambda {y^2} + 3x - 5y + 2 = 0$. Comparing this with $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a = 1$,$2h = -3$ (so $h = -\frac{3}{2}$),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 $\frac{1}{3} = \left| \frac{2\sqrt{(-\frac{3}{2})^2 - (1)(\lambda)}}{1 + \lambda} ight| = \frac{2\sqrt{\frac{9}{4} - \lambda}}{1 + \lambda}$.Squaring both sides: $\frac{1}{9} = \frac{4(\frac{9}{4} - \lambda)}{(1 + \lambda)^2} = \frac{9 - 4\lambda}{(1 + \lambda)^2}$.$(1 + \lambda)^2 = 9(9 - 4\lambda) \implies \lambda^2 + 2\lambda + 1 = 81 - 36\lambda$.$\lambda^2 + 38\lambda - 80 = 0$.Factoring the quadratic: $(\lambda + 40)(\lambda - 2) = 0$.Since $\lambda$ is a non-negative real number,we have $\lambda = 2$.

If the angle between the pair of straight lines represented by the equation ${x^2} - 3xy + \lambda {y^2} + 3x - 5y + 2 = 0$ is ${\tan ^{ - 1}}\left( {\frac{1}{3}} \right)$,where $\lambda$ is a non-negative real number,then $\lambda$ is:

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