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Given that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$, we can write $\vec{c} = -(\vec{a}+\vec{b})$.
Taking the magnitude squared of both sides, we have $|\vec{c}|^2 = |-(\vec{a}+\vec{b})|^2 = |\vec{a}+\vec{b}|^2$. Expanding the right side, we get $|\vec{c}|^2 = (\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b}$.
We are given $|\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7$. Substituting these values, we have $7^2 = 3^2 + 5^2 + 2\vec{a}\cdot\vec{b}$, which simplifies to $49 = 9 + 25 + 2\vec{a}\cdot\vec{b}$.
From the previous step, we have $49 = 34 + 2\vec{a}\cdot\vec{b}$, so $2\vec{a}\cdot\vec{b} = 49 - 34 = 15$. Thus, $\vec{a}\cdot\vec{b} = \frac{15}{2}$.
We know that $\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$. Substituting the given magnitudes and the calculated dot product, we have $\frac{15}{2} = (3)(5)\cos\theta$, which simplifies to $\frac{15}{2} = 15\cos\theta$.
Dividing both sides by 15, we get $\cos\theta = \frac{15}{2 \cdot 15} = \frac{1}{2}$. Therefore, $\theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$ or $60^\circ$.
Final Answer: $\frac{\pi}{3}$ or $60^\circ$
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