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(B) Given $A = \left(\frac{3}{\sqrt{a}}, \sqrt{a}\right)$.The image of $A$ in the $y$-axis is $B = \left(-\frac{3}{\sqrt{a}}, \sqrt{a}\right)$.The ima

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$8$

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(B) Given $A = \left(\frac{3}{\sqrt{a}}, \sqrt{a}ight)$.The image of $A$ in the $y$-axis is $B = \left(-\frac{3}{\sqrt{a}}, \sqrt{a}ight)$.The image of $B$ in the $x$-axis is $C = \left(-\frac{3}{\sqrt{a}}, -\sqrt{a}ight)$.The area of $	riangle ACD$ with vertices $A\left(\frac{3}{\sqrt{a}}, \sqrt{a}ight)$,$C\left(-\frac{3}{\sqrt{a}}, -\sqrt{a}ight)$,and $D(3 \cos 	heta, a \sin 	heta)$ is given by the determinant formula:$	ext{Area} = \frac{1}{2} \left| x_A(y_C - y_D) + x_C(y_D - y_A) + x_D(y_A - y_C) ight|$Substituting the coordinates:$	ext{Area} = \frac{1}{2} \left| \frac{3}{\sqrt{a}}(-\sqrt{a} - a \sin 	heta) - \frac{3}{\sqrt{a}}(a \sin 	heta - \sqrt{a}) + 3 \cos 	heta(\sqrt{a} - (-\sqrt{a})) ight|$$	ext{Area} = \frac{1}{2} \left| -3 - 3\sqrt{a} \sin 	heta - 3\sqrt{a} \sin 	heta + 3 + 6\sqrt{a} \cos 	heta ight|$$	ext{Area} = \frac{1}{2} \left| 6\sqrt{a} \cos 	heta - 6\sqrt{a} \sin 	heta ight| = 3\sqrt{a} |\cos 	heta - \sin 	heta|$Since $D$ is in the fourth quadrant,$	heta \in (\frac{3\pi}{2}, 2\pi)$,so $\cos 	heta > 0$ and $\sin 	heta  0$.The maximum value of $(\cos 	heta - \sin 	heta)$ is $\sqrt{1^2 + (-1)^2} = \sqrt{2}$.Therefore,$	ext{Max Area} = 3\sqrt{a} \cdot \sqrt{2} = 12$.$\sqrt{2a} = 4 \implies 2a = 16 \implies a = 8$.

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