If the angle between the vectors $\vec a$ and $\vec b$ is $\frac{π}{3}$ and the area of the triangle with adjacent sides parallel to $\vec a$ and $\vec b$ is 3, then $\vec a.\vec b$ is

$2\sqrt{3}$

We have,

$\frac{1}{2}|\vec a×\vec b|=3$

$⇒|\vec a×\vec b|=6⇒|\vec a||\vec b|\sin\frac{π}{3}=6⇒|\vec a||\vec b|=4\sqrt{3}$

$∴\vec a.\vec b=|\vec a||\vec b|\cos\frac{π}{3}=4\sqrt{3}×\frac{1}{2}=2\sqrt{3}$