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

Expanding $|\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 = 9$ with unit vectors gives $6 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 9$.
So $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -3/2$.
For the symmetric case where vectors are equally spaced (e.g., on a circle at 120° intervals), we have $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = -1/2$.
Expanding $|2\vec{a} + k\vec{b} + k\vec{c}|^2 = 9$ gives $4 + 2k^2 + 4k(\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}) + 2k^2\vec{b} \cdot \vec{c} = 9$.
Substituting the symmetric values: $4 + 2k^2(1/2) + 4k(-1) + 2k^2(-1/2) = 9$, which simplifies to $k^2 - 4k - 5 = 0$.
Factoring: $(k-5)(k+1) = 0$, giving $k = 5$ as the positive solution.