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(B) We have,$\cos \alpha = \frac{a \cdot b}{|a||b|}, \cos \beta = \frac{b \cdot c}{|b||c|}$ and $\cos \gamma = \frac{c \cdot a}{|c||a|}$.Given that $|

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$-\frac{3}{2}$

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(B) We have,$\cos \alpha = \frac{a \cdot b}{|a||b|}, \cos \beta = \frac{b \cdot c}{|b||c|}$ and $\cos \gamma = \frac{c \cdot a}{|c||a|}$.Given that $|a|=|b|=|c|=\lambda$ (where $\lambda > 0$),we have:$\cos \alpha = \frac{a \cdot b}{\lambda^2}, \cos \beta = \frac{b \cdot c}{\lambda^2}, \cos \gamma = \frac{c \cdot a}{\lambda^2}$.Therefore,$\cos \alpha + \cos \beta + \cos \gamma = \frac{a \cdot b + b \cdot c + c \cdot a}{\lambda^2} \quad \dots (i)$We know that $|a+b+c|^2 = (a+b+c) \cdot (a+b+c) = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a)$.Since $|a+b+c|^2 \geq 0$,we have:$|a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a) \geq 0$$\lambda^2 + \lambda^2 + \lambda^2 + 2(a \cdot b + b \cdot c + c \cdot a) \geq 0$$3\lambda^2 + 2(a \cdot b + b \cdot c + c \cdot a) \geq 0$$2(a \cdot b + b \cdot c + c \cdot a) \geq -3\lambda^2$$a \cdot b + b \cdot c + c \cdot a \geq -\frac{3}{2}\lambda^2 \quad \dots (ii)$Substituting $(ii)$ into $(i)$:$\cos \alpha + \cos \beta + \cos \gamma = \frac{a \cdot b + b \cdot c + c \cdot a}{\lambda^2} \geq \frac{-\frac{3}{2}\lambda^2}{\lambda^2} = -\frac{3}{2}$.Thus,the minimum value is $-\frac{3}{2}$.

If $a, b, c$ are vectors of equal magnitude such that $(a, b)=\alpha, (b, c)=\beta, (c, a)=\gamma$,then the minimum value of $\cos \alpha+\cos \beta+\cos \gamma$ is

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