If for theta in left[-frac{pi}{3}, 0right],th | vedclass

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(D) Given $x = 3 	an \left(	heta + \frac{\pi}{3}ight) = 3 \left( \frac{	an 	heta + \sqrt{3}}{1 - \sqrt{3} 	an 	heta} ight)$.Rearranging,$x(1

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$75$

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(D) Given $x = 3 	an \left(	heta + \frac{\pi}{3}ight) = 3 \left( \frac{	an 	heta + \sqrt{3}}{1 - \sqrt{3} 	an 	heta} ight)$.Rearranging,$x(1 - \sqrt{3} 	an 	heta) = 3 	an 	heta + 3\sqrt{3}$ $\Rightarrow x - 3\sqrt{3} = 	an 	heta (3 + \sqrt{3}x)$ $\Rightarrow 	an 	heta = \frac{x - 3\sqrt{3}}{3 + \sqrt{3}x} \dots (1)$.Given $y = 2 	an \left(	heta + \frac{\pi}{6}ight) = 2 \left( \frac{	an 	heta + \frac{1}{\sqrt{3}}}{1 - \frac{	an 	heta}{\sqrt{3}}} ight) = 2 \left( \frac{\sqrt{3} 	an 	heta + 1}{\sqrt{3} - 	an 	heta} ight)$.Rearranging,$y(\sqrt{3} - 	an 	heta) = 2\sqrt{3} 	an 	heta + 2 \dots (2)$.Substituting $(1)$ into $(2)$:$y \left( \sqrt{3} - \frac{x - 3\sqrt{3}}{3 + \sqrt{3}x} ight) = 2\sqrt{3} \left( \frac{x - 3\sqrt{3}}{3 + \sqrt{3}x} ight) + 2$.$y \left( \frac{3\sqrt{3} + 3x - x + 3\sqrt{3}}{3 + \sqrt{3}x} ight) = \frac{2\sqrt{3}x - 18 + 6 + 2\sqrt{3}x}{3 + \sqrt{3}x}$.$y(2x + 6\sqrt{3}) = 4\sqrt{3}x - 12 \Rightarrow 2xy + 6\sqrt{3}y = 4\sqrt{3}x - 12$.Dividing by $2$: $xy - 2\sqrt{3}x + 3\sqrt{3}y + 6 = 0$.Comparing with $xy + \alpha x + \beta y + \gamma = 0$,we get $\alpha = -2\sqrt{3}$,$\beta = 3\sqrt{3}$,$\gamma = 6$.Thus,$\alpha^2 + \beta^2 + \gamma^2 = (-2\sqrt{3})^2 + (3\sqrt{3})^2 + (6)^2 = 12 + 27 + 36 = 75$.

Column $I$	Column $II$
$(A)$ Circle	$(p)$ The locus of the point $(h, k)$ for which the line $h x+k y=1$ touches the circle $x^2+y^2=4$
$(B)$ Parabola	$(q)$ Points $z$ in the complex plane satisfying $\|z+2\|-\|z-2\|= \pm 3$
$(C)$ Ellipse	$(r)$ Points of the conic have parametric representation $x=\sqrt{3}\left(\frac{1-t^2}{1+t^2}\right), y=\frac{2 t}{1+t^2}$
$(D)$ Hyperbola	$(s)$ The eccentricity of the conic lies in the interval $1 \leq x < \infty$
	$(t)$ Points $z$ in the complex plane satisfying $\operatorname{Re}(z+1)^2=\|z\|^2+1$

If for $\theta \in \left[-\frac{\pi}{3}, 0\right]$,the points $(x, y) = \left(3 \tan \left(\theta+\frac{\pi}{3}\right), 2 \tan \left(\theta+\frac{\pi}{6}\right)\right)$ lie on $xy+\alpha x+\beta y+\gamma=0$,then $\alpha^2+\beta^2+\gamma^2$ is equal to:

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