Let A 3 a , a , a > 0 , be a fixed point in the x y -plane. The image of A in y -axis be B and the image…

$A=\left(\frac{3}{\sqrt{a}},\sqrt{a}\right)$

$B=\left(\frac{-3}{\sqrt{a}},\sqrt{a}\right)$ is the image of $A$ on y-axis

$C=\left(-\frac{3}{\sqrt{a}},-\sqrt{a}\right)$ is the image of $B$ in x-axis

$D\left(3\cos \theta ,a\sin \theta \right)$

Area of triangle $ACD$ will be

$\frac{1}{2}\left|\begin{array}{ccc}\frac{3}{\sqrt{a}}& \sqrt{a}& 1\\ -\frac{3}{\sqrt{a}}& -\sqrt{a}& 1\\ 3\cos \theta & a\sin \theta & 1\end{array}\right|$

$=3\sqrt{a}\left(\cos \theta -\sin \theta \right)$

The maximum area of triangle is $3\sqrt{a}\sqrt{2}$

i.e. $3\sqrt{a}\sqrt{2}=12$

$\Rightarrow a=8$