在$\triangle ABC$中，已知$\overrightarrow{AD} = 2\overrightarrow{DC}$，$AC = 3BC$，$\sin\angle BDC = 3\sin\angle BAC$，当$\overrightarrow{CA} \cdot \overrightarrow{CB} - \left| \overrightarrow{AB} \right|$取得最小值时，$\triangle ABC$的面积为（ ）
A．$\frac{\sqrt{3}}{2}$ B．$\frac{\sqrt{5}}{2}$ C．$\frac{3}{8}$ D．$\frac{3\sqrt{5}}{16}$
解：设$BC$，$AB$的边长分别为$a$，$c$，则$AD = 2DC = 2a$
在$\triangle ABD$中，$\frac{BD}{\sin\angle BAC} = \frac{AB}{\sin\angle BDA} = \frac{AB}{\sin\angle BDC} = \frac{AB}{3\sin\angle BAC}$
$\therefore BD = \frac{AB}{3} = \frac{c}{3}$
$\therefore$在$\triangle BDA$中，$\cos\angle BDA = \frac{\frac{c^{2}}{9} + 4a^{2} - c^{2}}{2 \cdot \frac{c}{3} \cdot 2a}$
在$\triangle BDC$中，$\cos\angle BDC = \frac{\frac{c^{2}}{9} + a^{2} - a^{2}}{2 \cdot \frac{c}{3} \cdot a}$
$\because \angle BDA + \angle BDC = \pi$，即$\cos\angle BDA = - \cos\angle BDC$
$\therefore \frac{\frac{c^{2}}{9} + 4a^{2} - c^{2}}{2 \cdot \frac{c}{3} \cdot 2a} = - \frac{\frac{c^{2}}{9} + a^{2} - a^{2}}{2 \cdot \frac{c}{3} \cdot a}$
$\therefore c = \sqrt{6}a$
$\therefore$在$\triangle BCA$中，$\cos C = \frac{a^{2} + 9a^{2} - 6a^{2}}{2 \cdot a \cdot 3a} = \frac{2}{3}$
$\therefore \overrightarrow{CA} \cdot \overrightarrow{CB} - \left| \overrightarrow{AB} \right| = a \cdot 3a \cdot \frac{2}{3} - \sqrt{6}a = 2a^{2} - \sqrt{6}a = 2\left( a - \frac{\sqrt{6}}{4} \right)^{2} - \frac{3}{4}$
$\therefore a = \frac{\sqrt{6}}{4}$时，$\overrightarrow{CA} \cdot \overrightarrow{CB} - \left| \overrightarrow{AB} \right|$最小
$\therefore S_{\triangle ABC} = \frac{1}{2} \cdot a \cdot 3a \cdot \sin C = \frac{1}{2} \cdot 3 \cdot \frac{3}{8} \cdot \frac{\sqrt{5}}{3} = \frac{3\sqrt{5}}{16}$，故选 D