Let  $\mathrm{A}\left(\frac{3}{\sqrt{\mathrm{a}}},\sqrt{\mathrm{a}}\right)\mathrm{a}>0$  be a fixed point in the xy-plane. The image of A in y-axis be B and the image of B in x-axis be C. If D(3 cos $\mathrm{\theta }$, a sin $\mathrm{\theta }$) is a point in the fourth quadrant such that the maximum area of$\mathrm{\Delta }$ACD is 12 square units, then a is equal to _______

$\begin{array}{l}\mathrm{A}=\left(\frac{3}{\sqrt{\mathrm{a}}},\sqrt{\mathrm{a}}\right)\mathrm{The} \mathrm{image} \mathrm{of} \mathrm{A} \mathrm{in} \mathrm{y}-\mathrm{axis} \mathrm{be} \mathrm{B}\\ \Rightarrow \mathrm{B}=\left(\frac{-3}{\sqrt{\mathrm{a}}},\sqrt{\mathrm{a}}\right)\mathrm{The} \mathrm{image} \mathrm{of} \mathrm{B} \mathrm{in} \mathrm{x}-\mathrm{axis} \mathrm{be} \mathrm{C}\\ \\ \Rightarrow \mathrm{C}=\left(-\frac{3}{\sqrt{\mathrm{a}}},-\sqrt{\mathrm{a}}\right)\end{array}$
$\begin{array}{l}\text{ Area of }\mathrm{△}ACD=\frac{1}{2}\left|\begin{array}{ccc}\frac{3}{\sqrt{a}}& \sqrt{a}& 1\\ \frac{-3}{\sqrt{a}}& -\sqrt{a}& 1\\ 3\cos \theta & a\sin \theta & 1\end{array}\right|\\ =\frac{1}{2}\left(\frac{3}{\sqrt{a}}(-\sqrt{a}-a\sin \theta )-\sqrt{a}\left(\frac{-3}{\sqrt{a}}\cdot 1-3\cos \theta \right)+1\left(\frac{-3}{\sqrt{a}}\cdot a\sin \theta +3\sqrt{a}\cdot \cos \theta \right)\right)\\ =\frac{1}{2}(-3-3\sqrt{a}\sin \theta +3+3\sqrt{a}\cos \theta -3\sqrt{a}\sin \theta +3\sqrt{a}\cos \theta )\\ =\frac{1}{2}(-6\sqrt{a}\sin \theta +6\sqrt{a}\cos \theta )\\ =\frac{6}{2}\sqrt{a}(\cos \theta -\sin \theta )\\ =3\sqrt{a}(\cos \theta -\sin \theta )\end{array}$

max values of function is $3\sqrt{\mathrm{a}}\sqrt{2}$
$\begin{array}{l}3\sqrt{\mathrm{a}}\sqrt{2}=12\\ 2\mathrm{a}=16\\ \mathrm{a}=8\end{array}$