Prove area relations of triangle with giving vector equation 
In a triangle $ABC$ with point $I$ inside it such that $\overrightarrow{IA}+2\overrightarrow{IB}+3\overrightarrow{IC}=0$. If $S$ is the area of $\triangle ABC$ and $T$ is the area of $\triangle AIC$ then prove that $S=3T$

By the given we could get:
\begin{equation}
-\overrightarrow{AI}+2\overrightarrow{IB}+3\overrightarrow{IC}=0\\-6\overrightarrow{AI}+2\overrightarrow{AB}+3\overrightarrow{AC}=0\\\Rightarrow 6\overrightarrow{IA}+2\overrightarrow{AB}+3\overrightarrow{AC}=0
\end{equation}
This is all I could get from the given, moreover I don't know if this problem is in 2D or 3D at all but I think maybe it is in 2D. The problem ask to prove about area so we should find some relations of the cross product that I couldn't. And for the picture I don't know how to draw it too. Please help
 A: Notice that $$\overrightarrow{IA}+2\overrightarrow{IB}+3\overrightarrow{IC}=0\iff 3\cdot \frac{\overrightarrow{IA}+2\overrightarrow{IB}}{3}+3\overrightarrow{IC}=\overrightarrow 0$$ Thus, consider the point $P$ on the plane such that $$\vec{p}-\vec{i}=\overrightarrow{IP}=\frac{\overrightarrow{IA}+2\overrightarrow{IB}}{3}=\frac{\left(\vec{a}-\vec{i}\right)+2\cdot \left(\vec{b}-\vec{i}\right)}{3}=\frac{\vec{a}+2\vec{b}}{3}-\vec{i}\iff \vec{p}=\frac{\vec{a}+2\vec{b}}3$$ Hence, $P$ is the point on $AB$ such that $AP=2PB$. Go back to the original equation and substitute in $$3\overrightarrow{IP}+3\overrightarrow{IC}=0\iff \overrightarrow{IP}=\overrightarrow{CI}$$ Thus, $I$ is the midpoint of $CP$. Finally, we have $$\fbox{$\frac{S}{T}=\frac{S}{\frac12\triangle PAC}=\frac{S}{\frac12\cdot \frac23 S}=3$}$$
A: First, yes, it is planar, because a linear combination of $\overrightarrow{IA}, \overrightarrow{IB},$ and $\overrightarrow{IC} = 0$, which means that you can express any combination of them with only two, hence two dimensional.
You made a good start.  Remember we are doing a comparison.  The area of $\Delta ABC$ is $\frac12 \lvert \overrightarrow{AC} \times  \overrightarrow{AB}\rvert$ and the area of $\Delta AIC$ is $\frac12 \lvert \overrightarrow{AC} \times \overrightarrow{AI}\rvert$, so we want to rewrite one in terms of the other somehow.  Solve your expressions for $\overrightarrow{AI} = \frac13 \overrightarrow{AB} + \frac12\overrightarrow{AC}$.  Then $\overrightarrow{AC} \times \overrightarrow{AB} = \overrightarrow{AC} \times\left(\frac13\overrightarrow{AB} + \frac12 \overrightarrow{AC}\right)$ and $\overrightarrow{AC} \times \overrightarrow{AC} = \overrightarrow{0}$. The result follows.
