If $A(a)$, $B(b)$, and $C(c)$ are vertices of $\Delta ABC$. Point $D$ divides segment $BC$ internally in the ratio $2 : 1$. Point $E$ divides segment $AD$ internally in the ratio $1 : 2$, then the position vector of $E$ is
- A$\frac{3a + 2b + c}{6}$
- B$\frac{6a + 2b + c}{9}$
- C$\frac{3a + 2b + 4c}{9}$
- D$\frac{a + 2b + c}{3}$
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