Column I Column II ( A ) In ℝ ^ 2 , if the magnitude of the...

\left(A\right)$As \left|\dfrac{\sqrt{3}\alpha + \beta}{2}\right| = \sqrt{3} ⇒ \left|\sqrt{3}\alpha + \beta\right| = 2\sqrt{3}$ ...(i)
And, also $\alpha = 2 + \sqrt{3}\beta$ ...(ii)
Solving (i) & (ii), we get
$⇒ 2\sqrt{3} + 4\beta = \pm 2\sqrt{3}$
$⇒ \beta = 0, - \sqrt{3}$
So, $\alpha = 2$ and $- 1$
i.e, \left|\alpha\right| = 2, 1$<br />$( B )$Here$- 3a - 2 = b + {a}^{2}$...(i)<br />And$- 6a = b$...(ii)<br />Solving (i) and (ii),<br />$- 3a - 2 = - 6a + {a}^{2}$<br />$⇒ {a}^{2} - 3a + 2 = 0$<br />$⇒ a = 1, 2$<br />\left(C\right)$ Here, ${\left(3 - 3\omega + 2{\omega}^{2}\right)}^{4n + 3}\left\{1 + {\omega}^{4n + 3} + {\left({\omega}^{2}\right)}^{4n + 3}\right\}$$<br />$⇒ {\left(3 - 3\omega + 2{\omega}^{2}\right)}^{4n + 3}\left\{1 + {\omega}^{4n} + {\omega}^{8n}\right\} = 0$<br />So,$n = 1, 2, 4, 5$<br />\left(D\right)$q - a = 2d$<br />Let$q - q = \alpha, d = \dfrac{\alpha}{2}$<br />$a = 5 - \dfrac{\alpha}{2}, b = a + \dfrac{3\alpha}{2} = 5 + \alpha$<br />Now,$\dfrac{ab}{a + b} = 2$<br />$⇒ \dfrac{\left(5 - \dfrac{\alpha}{2}\right)\left(5 + \alpha\right)}{5 - \dfrac{\alpha}{2} + 5 + \alpha} = 2$<br />$⇒ \alpha = 5, - 2$<br />\left|\alpha\right| = 5, 2$