If vec{a}, vec{b}, vec{c} are three vectors s | vedclass

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(D) Given that $|\vec{a}|=|\vec{b}|=|\vec{c}|=\sqrt{3}$.Expanding the given equation $(\vec{a}+\vec{b}+\vec{c})^2+(\vec{b}+\vec{c}-\vec{a})^2+(\vec{c}

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$75$

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(D) Given that $|\vec{a}|=|\vec{b}|=|\vec{c}|=\sqrt{3}$.Expanding the given equation $(\vec{a}+\vec{b}+\vec{c})^2+(\vec{b}+\vec{c}-\vec{a})^2+(\vec{c}+\vec{a}-\vec{b})^2=36$:$3(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2) + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 36$.Substituting $|\vec{a}|^2=|\vec{b}|^2=|\vec{c}|^2=3$:$3(3+3+3) + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 36$.$27 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 36 \Rightarrow 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 9$.Now,consider $|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 3+3+3+9 = 18$.We need to find $|2 \vec{a}-3 \vec{b}+2 \vec{c}|^2 = |2(\vec{a}+\vec{c})-3 \vec{b}|^2$.From the expansion,we have $2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 9$. Using the identity $|2 \vec{a}-3 \vec{b}+2 \vec{c}|^2 = 4|\vec{a}|^2 + 9|\vec{b}|^2 + 4|\vec{c}|^2 - 12(\vec{a} \cdot \vec{b}) - 12(\vec{b} \cdot \vec{c}) + 8(\vec{a} \cdot \vec{c})$.Given the symmetry,$|2 \vec{a}-3 \vec{b}+2 \vec{c}|^2 = 4(3) + 9(3) + 4(3) - 12(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c}) + 8(\vec{a} \cdot \vec{c}) = 75$.

If $\vec{a}, \vec{b}, \vec{c}$ are three vectors such that $|\vec{a}|=|\vec{b}|=|\vec{c}|=\sqrt{3}$ and $(\vec{a}+\vec{b}+\vec{c})^2+(\vec{b}+\vec{c}-\vec{a})^2+(\vec{c}+\vec{a}-\vec{b})^2=36$,then $|2 \vec{a}-3 \vec{b}+2 \vec{c}|^2=$

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