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

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$25$

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(B) Given that $|\vec{u}+\vec{v}+\vec{w}|=0$. Squaring both sides,we get $|\vec{u}+\vec{v}+\vec{w}|^2 = 0^2$. Using the identity $|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})$,we have: $|\vec{u}|^2+|\vec{v}|^2+|\vec{w}|^2+2(\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}) = 0$. Substituting the given values $|\vec{u}|=3, |\vec{v}|=4, |\vec{w}|=5$: $3^2+4^2+5^2+2(\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}) = 0$. $9+16+25+2(\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}) = 0$. $50+2(\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}) = 0$. $2(\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}) = -50$. $\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u} = -25$. Taking the absolute value as requested: $|\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}| = |-25| = 25$.

Let $\vec{u}, \vec{v}$ and $\vec{w}$ be vectors such that $|\vec{u}+\vec{v}+\vec{w}|=0$. If $|\vec{u}|=3$,$|\vec{v}|=4$ and $|\vec{w}|=5$,then the value of $|\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}|$ is

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