Given: \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\), \(|\vec{a}| = \sqrt{37}\), \(|\vec{b}| = 3\), \(|\vec{c}| = 4\)

We can rewrite the given equation as: \(\vec{a} = -(\vec{b} + \vec{c})\)

Taking the magnitude squared of both sides: \(|\vec{a}|^2 = |-(\vec{b} + \vec{c})|^2\)

\(|\vec{a}|^2 = |\vec{b} + \vec{c}|^2\)

Expanding the right side using the dot product: \(|\vec{a}|^2 = (\vec{b} + \vec{c}) \cdot (\vec{b} + \vec{c})\)

\(|\vec{a}|^2 = \vec{b} \cdot \vec{b} + 2(\vec{b} \cdot \vec{c}) + \vec{c} \cdot \vec{c}\)

\(|\vec{a}|^2 = |\vec{b}|^2 + 2(\vec{b} \cdot \vec{c}) + |\vec{c}|^2\)

We know that \(\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos{\theta}\), where \(\theta\) is the angle between \(\vec{b}\) and \(\vec{c}\).

So, \(|\vec{a}|^2 = |\vec{b}|^2 + 2|\vec{b}| |\vec{c}| \cos{\theta} + |\vec{c}|^2\)

Substituting the given values: \((\sqrt{37})^2 = (3)^2 + 2(3)(4) \cos{\theta} + (4)^2\)

\(37 = 9 + 24 \cos{\theta} + 16\)

\(37 = 25 + 24 \cos{\theta}\)

\(12 = 24 \cos{\theta}\)

\(\cos{\theta} = \dfrac{12}{24} = \dfrac{1}{2}\)

Therefore, \(\theta = \arccos{\left(\dfrac{1}{2}\right)}\)

\(\theta = \dfrac{\pi}{3}\)