Let A( 3 a , a ), a > 0 , be a fixed point in the xy -plane. The...

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

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

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

$D\left(3cos\theta, asin\theta\right)$

Area of triangle $ACD$ will be

\dfrac{1}{2}\left|\begin{cases}\dfrac{3}{\sqrt{a}} & \sqrt{a} & 1 \\ - \dfrac{3}{\sqrt{a}} & - \sqrt{a} & 1 \\ 3cos\theta & asin\theta & 1\end{cases}\right|

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

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

i.e. $3\sqrt{a}\sqrt{2} = 12$

$⇒ a = 8$