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(B) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,we have $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1.$We are given the equation $\vec{a}+\vec{b}

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$-3/2$

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(B) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,we have $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1.$We are given the equation $\vec{a}+\vec{b}+\vec{c}=\vec{0}.$Squaring both sides,we get:$|\vec{a}+\vec{b}+\vec{c}|^{2} = |\vec{0}|^{2}$Expanding the dot product:$(\vec{a}+\vec{b}+\vec{c}) \cdot (\vec{a}+\vec{b}+\vec{c}) = 0$$|\vec{a}|^{2} + |\vec{b}|^{2} + |\vec{c}|^{2} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$Substituting the magnitudes $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$:$1^{2} + 1^{2} + 1^{2} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$$3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$Solving for the required expression:$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -3$$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}$

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\vec{0},$ find the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}.$

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