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(A) The given equation is $3x^2 + axy - 2y^2 = 0$. Dividing by $x^2$,we get $3 + a(\frac{y}{x}) - 2(\frac{y}{x})^2 = 0$. Let $m = \frac{y}{x}$ be the

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$\sqrt{3}$

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(A) The given equation is $3x^2 + axy - 2y^2 = 0$. Dividing by $x^2$,we get $3 + a(\frac{y}{x}) - 2(\frac{y}{x})^2 = 0$. Let $m = \frac{y}{x}$ be the slope of the lines. Then $2m^2 - am - 3 = 0$. Since one line makes an angle of $60^{\circ}$ with the $x$-axis,its slope is $m = 	an(60^{\circ}) = \sqrt{3}$. Substituting $m = \sqrt{3}$ into the quadratic equation: $2(\sqrt{3})^2 - a(\sqrt{3}) - 3 = 0$ $2(3) - a\sqrt{3} - 3 = 0$ $6 - 3 = a\sqrt{3}$ $3 = a\sqrt{3}$ $a = \frac{3}{\sqrt{3}} = \sqrt{3}$.

If one of the lines given by the pair of lines $3x^2 + axy - 2y^2 = 0$ makes an angle of $60^{\circ}$ with the $x$-axis,then $a=$

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