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(B) Given, $\vec{a}+\vec{b}+\vec{c}=\vec{0}$.Squaring both sides, we get $|\vec{a}+\vec{b}+\vec{c}|^2 = 0$.Using the identity $|\vec{a}+\vec{b}+\vec{c

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$\frac{7}{2}$

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(B) Given, $\vec{a}+\vec{b}+\vec{c}=\vec{0}$.Squaring both sides, we get $|\vec{a}+\vec{b}+\vec{c}|^2 = 0$.Using the identity $|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) = 0$.Substituting the given magnitudes $|\vec{a}|=3, |\vec{b}|=4, |\vec{c}|=2$:$3^2+4^2+2^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) = 0$.$9+16+4+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) = 0$.$29+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) = 0$.$\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a} = -\frac{29}{2}$.Now, we need to find $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}+2(|\vec{a}|+|\vec{b}|+|\vec{c}|)$.$= -\frac{29}{2} + 2(3+4+2) = -\frac{29}{2} + 2(9) = -\frac{29}{2} + 18 = \frac{-29+36}{2} = \frac{7}{2}$.

Three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ satisfy $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. If $|\vec{a}|=3, |\vec{b}|=4, |\vec{c}|=2$, then $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}+2(|\vec{a}|+|\vec{b}|+|\vec{c}|)=$

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