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

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(C) Given that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.Taking the dot product of the sum with itself: $(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} +

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$-7$

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(C) Given that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.Taking the dot product of the sum with itself: $(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = \vec{0} \cdot \vec{0}$.This expands to $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.Substituting the given magnitudes $|\vec{a}| = 1, |\vec{b}| = 2, |\vec{c}| = 3$:$1^2 + 2^2 + 3^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.$1 + 4 + 9 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.$14 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -14$.Therefore,$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -7$.

If $\vec{a}, \vec{b}, \vec{c}$ are three vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$,$|\vec{a}| = 1$,$|\vec{b}| = 2$,and $|\vec{c}| = 3$,then find the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$.

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