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(D) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$. Expanding the given equation: $|\vec{a}-\vec{

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$5$

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(D) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$. Expanding the given equation: $|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9$ $(|\vec{a}|^{2}+|\vec{b}|^{2}-2\vec{a}\cdot\vec{b}) + (|\vec{b}|^{2}+|\vec{c}|^{2}-2\vec{b}\cdot\vec{c}) + (|\vec{c}|^{2}+|\vec{a}|^{2}-2\vec{c}\cdot\vec{a}) = 9$ $2(|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}) - 2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) = 9$ $2(1+1+1) - 2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) = 9$ $6 - 2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) = 9$ $\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a} = -\frac{3}{2}$ Now, consider $|\vec{a}+\vec{b}+\vec{c}|^{2} = |\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2} + 2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) = 3 + 2(-\frac{3}{2}) = 0$. Thus, $\vec{a}+\vec{b}+\vec{c} = 0$, which implies $\vec{b}+\vec{c} = -\vec{a}$. Substitute this into the second equation: $|2\vec{a}+k(\vec{b}+\vec{c})| = 3$ $|2\vec{a}+k(-\vec{a})| = 3$ $|(2-k)\vec{a}| = 3$ Since $|\vec{a}| = 1$, we have $|2-k| = 3$. This gives $2-k = 3$ or $2-k = -3$. $k = -1$ or $k = 5$. The positive value of $k$ is $5$.

For three unit vectors $\vec{a}, \vec{b}, \vec{c}$ satisfying $|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9$ and $|2\vec{a}+k\vec{b}+k\vec{c}|=3$, the positive value of $k$ is:

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