Explanation

(d)
$A \equiv (a\cos\theta,\ b\sin\theta)$
$B \equiv (a\cos{(\theta + \alpha)},\ b\sin{(\theta + \alpha)})$
$C \equiv (a\cos{(\theta + 2\alpha)},\ b\sin{(\theta + 2\alpha)})$
$\mathrm{\Delta}\ \equiv Area\ of\ \mathrm{\Delta}ABC$
$= \frac{1}{2}\left| \begin{matrix}
1 & a\cos\theta & b\sin\theta \\
1 & a\cos{(\theta + \alpha)} & b\sin{(\theta + \alpha)} \\
1 & a\cos{(\theta + 2\alpha)} & b\sin{(\theta + 2\alpha)}
\end{matrix} \right|$
$= 2ab\sin^{2}\left( \frac{\alpha}{2} \right)\sin\alpha$
$\mathrm{\Delta}(\alpha) = ab\sin{\alpha(1 - \cos\alpha)}$
$= \frac{ab}{2}\left( 2\sin\alpha - \sin{2\alpha} \right)$
$\mathrm{\Delta}^{'}(\alpha) = 0$
$\Rightarrow \cos\alpha = 1$
Or $\cos\alpha = - \frac{1}{2}$
$\cos\alpha = 1$ gives $\mathrm{\Delta} = 0$
$\cos\alpha = - \frac{1}{2}$ gives maximum value of $\mathrm{\Delta} = \frac{3\sqrt{3}}{4}ab$