If vec a, vec b and vec c be vectors such that vec a+vec b+vec c=vec…

The correct answer is Option (2) → $\frac{\pi}{3}$

Given: $\vec{a} + \vec{b} + \vec{c} = \vec{0} \Rightarrow \vec{c} = -(\vec{a} + \vec{b})$

Take magnitudes squared:

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

$\Rightarrow c^{2} = a^{2} + b^{2} + 2ab\cos\theta$

Substitute $a=3,\;b=5,\;c=7$

$7^{2} = 3^{2} + 5^{2} + 2(3)(5)\cos\theta$

$49 = 9 + 25 + 30\cos\theta$

$49 - 34 = 30\cos\theta$

$15 = 30\cos\theta$

$\cos\theta = \frac{1}{2}$

$\Rightarrow \theta = 60^{\circ}$

Angle between $\vec{a}$ and $\vec{b}$ is $60^{\circ}$