Let D and E be the midpoints of the sides AC | vedclass

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(D) Let the position vectors of $A, B, C$ with respect to the origin $O$ be $\vec{a}, \vec{b}, \vec{c}$ respectively.Given that $\vec{a}+2\vec{b}+3\ve

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(D) Let the position vectors of $A, B, C$ with respect to the origin $O$ be $\vec{a}, \vec{b}, \vec{c}$ respectively.Given that $\vec{a}+2\vec{b}+3\vec{c}=\vec{0}$.Since $D$ is the midpoint of $AC$,the position vector of $D$ is $\vec{d} = \frac{\vec{a}+\vec{c}}{2}$.Since $E$ is the midpoint of $BC$,the position vector of $E$ is $\vec{e} = \frac{\vec{b}+\vec{c}}{2}$.The area of $	riangle ODE$ is given by $	ext{Area} = \frac{1}{2} |\vec{d} 	imes \vec{e}|$.Substituting the values of $\vec{d}$ and $\vec{e}$:$	ext{Area} = \frac{1}{2} |(\frac{\vec{a}+\vec{c}}{2}) 	imes (\frac{\vec{b}+\vec{c}}{2})| = \frac{1}{8} |\vec{a} 	imes \vec{b} + \vec{a} 	imes \vec{c} + \vec{c} 	imes \vec{b} + \vec{c} 	imes \vec{c}|$.Since $\vec{c} 	imes \vec{c} = \vec{0}$,we have $	ext{Area} = \frac{1}{8} |\vec{a} 	imes \vec{b} + \vec{a} 	imes \vec{c} - \vec{b} 	imes \vec{c}|$.From the given equation $\vec{a}+2\vec{b}+3\vec{c}=\vec{0}$,taking the cross product with $\vec{b}$ gives $\vec{a} 	imes \vec{b} + 3(\vec{c} 	imes \vec{b}) = \vec{0} \Rightarrow \vec{a} 	imes \vec{b} = 3(\vec{b} 	imes \vec{c})$.Taking the cross product with $\vec{c}$ gives $\vec{a} 	imes \vec{c} + 2(\vec{b} 	imes \vec{c}) = \vec{0} \Rightarrow \vec{a} 	imes \vec{c} = 2(\vec{c} 	imes \vec{b})$.Substituting these into the area expression:$	ext{Area} = \frac{1}{8} |3(\vec{b} 	imes \vec{c}) + 2(\vec{c} 	imes \vec{b}) - (\vec{b} 	imes \vec{c})| = \frac{1}{8} |3(\vec{b} 	imes \vec{c}) - 2(\vec{b} 	imes \vec{c}) - (\vec{b} 	imes \vec{c})| = \frac{1}{8} |0| = 0$.

Let $D$ and $E$ be the midpoints of the sides $AC$ and $BC$ of a triangle $ABC$ respectively. If $O$ is an interior point of the triangle $ABC$ such that $\overrightarrow{OA}+2\overrightarrow{OB}+3\overrightarrow{OC}=\overrightarrow{0}$,then the area (in sq. units) of the triangle $ODE$ is

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