If A(a), B(b), and C(c) are vertices of Delta | vedclass

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(D) Step $1$: Find the position vector of point $D$ which divides $BC$ in ratio $2:1$. Using section formula, $\vec{d} = \frac{2\vec{c} + 1\vec{b}}{2+

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$\frac{a + 2b + c}{3}$

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(D) Step $1$: Find the position vector of point $D$ which divides $BC$ in ratio $2:1$. Using section formula, $\vec{d} = \frac{2\vec{c} + 1\vec{b}}{2+1} = \frac{\vec{b} + 2\vec{c}}{3}$.Step $2$: Point $E$ divides $AD$ in ratio $1:2$. Using section formula, $\vec{e} = \frac{1(\vec{d}) + 2(\vec{a})}{1+2} = \frac{\vec{d} + 2\vec{a}}{3}$.Step $3$: Substitute $\vec{d}$ into the expression for $\vec{e}$: $\vec{e} = \frac{\frac{\vec{b} + 2\vec{c}}{3} + 2\vec{a}}{3} = \frac{\vec{b} + 2\vec{c} + 6\vec{a}}{9} = \frac{6\vec{a} + \vec{b} + 2\vec{c}}{9}$.

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

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