If for $\theta \in \left[-\frac{\pi }{3},0\right]$, the points $(\mathrm{x},\mathrm{y})=\left(3\tan \left(\theta +\frac{\pi }{3}\right),2\tan \left(\theta +\frac{\pi }{6}\right)\right)$ lie on $xy+\alpha x+\beta y+\gamma =0$, then ${\alpha }^{2}+{\beta }^{2}+{\gamma }^{2}$ is equal to : [JEE Main 2025, 7 Apr (Shift 1)]

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YESPYQ · Accepted Answer

Correct answer: d) $75$. Given $(\mathrm{x},\mathrm{y})=\left(3\tan \left(\theta +\frac{\pi }{3}\right),2\tan \left(\theta +\frac{\pi }{6}\right)\right)$ $ \mathrm{x}=3\left(\frac{\tan \theta+\sqrt{3}}{1-\sqrt{3} \tan \theta}\right) $ $ \mathrm{x}-\sqrt{3} \tan \theta=3 \tan \theta+3 \sqrt{3} $ $ \tan \theta=\frac{x-3 \sqrt{3}}{3+\sqrt{3} x} \quad \ldots(1)$ $ 2\left(\frac{\tan \theta+\frac{1}{\sqrt{3}}}{1-\frac{\tan \theta}{\sqrt{3}}}=y\right) $ $ 2(\sqrt{3} \tan \theta+1)=y(\sqrt{3}-\tan \theta) \ldots(2) $ using (1) and (2) $ 2\left(\frac{x-3 \sqrt{3}}{\sqrt{3}+x}+1\right)=y\left(\sqrt{3}-\frac{(x-3 \sqrt{3})}{\sqrt{3}(\sqrt{3}+x)}\right) $ $ 2 \sqrt{3}(x-3 \sqrt{3}+x+\sqrt{3})=y(3(\sqrt{3}+x)-x+3 \sqrt{3})$ $ 4 \sqrt{3} x-12=y(2 x+6 \sqrt{3}) $ $x y-2 \sqrt{3} x+3 \sqrt{3} y-6=0 $ $ \Rightarrow \alpha=-2 \sqrt{3}, \beta=3 \sqrt{3}, \gamma=-6$ $ \alpha^2+\beta^2+\gamma^2=12+27+36=75$

If for \(\theta \in \left[-\frac{\pi }{3},0\right]\), the points \((\mathrm{x},\mathrm{y})=\left(3\tan \left(\theta +\fr…

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