How find this value $\frac{S_{\Delta ABC}}{\overrightarrow{OA}\cdot\overrightarrow{OB}}$ 
In the plane have $4$ point $O,A,B,C$ such
  $$\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{0},2|\overrightarrow{OA}|=2|\overrightarrow{OB}|+\overrightarrow{OC}$$,and let
  $$\theta=<\overrightarrow{AB},\overrightarrow{OC}>,\tan{\theta}=\dfrac{3}{2}$$

Find the value $$\dfrac{S_{\Delta ABC}}{\overrightarrow{OA}\cdot\overrightarrow{OB}}$$

My try: since 
$$\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{0}$$
then $O$ is The center of gravity with $\Delta ABC$.
so 
$$2|\overrightarrow{OA}|=2|\overrightarrow{OB}|+\overrightarrow{OC}\Longrightarrow \dfrac{4}{3}|AD|=\dfrac{4}{3}|BE|+\dfrac{2}{3}|CF|$$
$$\Longrightarrow |AD|=|BE|+2|CF|$$
then I can't.Thank you

The answer as given by
orion
is repeated here for convenience.
This answer is certainly incomplete, but let it be assumed that it is correct so far.
$$
c^2+(a-b)^2-2c(a-b)\cos\theta=a^2 \\
c^2+(a-b)^2+2c(a-b)\cos\theta=b^2 \\
(2c)^2=a^2+b^2-2ab\cos\alpha
$$
Quote: "but $a,b,c$ can be scaled without affecting the outcome".
Hence, without loss of generality we can put (WLOG equation):
$$
a^2+b^2+c^2 = 1
$$
Orion's equations can be simplified a great deal.
Subtracting the first two and using $\;\tan{\theta}=3/2$ :
$$
-4c(a-b)\cos\theta=a^2-b^2 \quad \Longrightarrow \quad -4c\cos\theta=a+b
\quad \Longrightarrow \quad a+b=-4c\frac{2}{\sqrt{13}}
$$
Adding the first two equations:
$$
2c^2+2(a-b)^2=a^2+b^2 \quad \Longrightarrow \quad c^2+(a^2+b^2+c^2)-4ab=0
\quad \Longrightarrow \quad c^2-4ab=-1
$$
And orion's last equation:
$$
(2c)^2+c^2=(a^2+b^2+c^2)-2ab\,\cos\alpha \quad \Longrightarrow \quad 5c^2+2\cos\alpha \,ab=1
$$
Hence we can solve $c^2$ and $ab$ from:
$$
\left[ \begin{array}{cc} 1 & -4 \\ 5 & 2\cos\alpha \end{array} \right]
\left[ \begin{array}{c} c^2 \\ ab \end{array} \right] =
\left[ \begin{array}{c} -1 \\ 1 \end{array} \right] \quad \Longrightarrow \quad
\left[ \begin{array}{c} c^2 \\ ab \end{array} \right] = 
\frac{1}{10+\cos\alpha} \left[ \begin{array}{c} 2-\cos\alpha \\ 3 \end{array} \right]
$$
We also have:
$$
a+b=-4c\frac{2}{\sqrt{13}} \qquad \Longrightarrow \qquad (a+b)^2 = \frac{64}{13}\frac{2-\cos\alpha}{10+\cos\alpha}
$$
Hence our WLOG equation can be rewritten as:
$$
a^2+b^2+c^2 = (a+b)^2 - 2ab + c^2 = 1 \qquad \Longrightarrow \\
\frac{64}{13}(2-\cos\alpha)-2\cdot3+(2-\cos\alpha)=10+\cos\alpha \quad \Longrightarrow \quad \cos\alpha = -\frac{3}{5}
$$
Now $(3,4,5)$ is a famous rectangular triangle. So what we finally get is an incredibly simple answer:
$$ \frac{S}{\vec{OA}\cdot\vec{OB}}=\frac32\tan \alpha \quad \wedge \quad 
\tan\alpha = -\frac{4}{3} \qquad \Longrightarrow \qquad
\frac{S}{\vec{OA}\cdot\vec{OB}}= -2
$$
Where the minus sign must be due to the fact that the angle AOB is obtuse.

Bonus. Calculating $\;a,b,c\;$ too.
$$
c = \pm \sqrt{\frac{2-\cos\alpha}{10+\cos\alpha}} = \pm \sqrt{\frac{13}{47}} \\
a+b = -\frac{8c}{\sqrt{13}} > 0 \quad \Longrightarrow \quad \frac{a+b}{2} = \frac{4}{\sqrt{47}} \qquad ; \qquad ab = \frac{3}{10+\cos\alpha}=\frac{15}{47} \\ \Longrightarrow \qquad
\{a,b\} = \frac{a+b}{2} \pm \sqrt{\left(\frac{a+b}{2}\right)^2-ab} \; = \; \frac{4 \pm 1}{\sqrt{47}}
$$
Without the WLOG restriction (or $\;a^2+b^2+c^2=47\,t^2$ ) with $t$ an arbitrary positive real constant:
$$
c = \sqrt{13}\,t \quad ; \quad a = 5\,t \quad ; \quad b = 3\,t
$$
Where it is noted that $\,a\,$ and $\,b\,$ can be swapped.
