A,B,C,D are four points in the space and satisfy $\mid \overrightarrow{AB}\mid =3,\mid \overrightarrow{BC}\mid =7,\mid \overrightarrow{CD}\mid =11$ and $\mid \overrightarrow{DA}\mid =9$. Then $\overrightarrow{AC}.\overrightarrow{BD}$ is

I know that $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AD}=0$ and also I observed that

$\mid \overrightarrow{AB}{\mid}^{2}+\mid \overrightarrow{CD}{\mid}^{2}=\mid \overrightarrow{BC}{\mid}^{2}+\mid \overrightarrow{DA}{\mid}^{2}$. But couldn't get anything from it. Any hint?

I know that $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AD}=0$ and also I observed that

$\mid \overrightarrow{AB}{\mid}^{2}+\mid \overrightarrow{CD}{\mid}^{2}=\mid \overrightarrow{BC}{\mid}^{2}+\mid \overrightarrow{DA}{\mid}^{2}$. But couldn't get anything from it. Any hint?