Solution:
Given,
$|\vec{\text{a}}|=3,\big|\vec{\text{b}}\big|=5$ and $|\vec{\text{c}}|=7\dots(1)$Let
$\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$Given that
$\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=0$
$\Rightarrow\vec{\text{a}}+\vec{\text{b}}=-\vec{\text{c}}$
$\Rightarrow\big|\vec{\text{a}}+\vec{\text{b}}\big|=|-\vec{\text{c}}|^2$
$\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{c}}|^2$
$\Rightarrow2\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{c}}|^2-|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2$
$\Rightarrow2\vec{\text{a}}.\vec{\text{b}}=7^2-3^3-5^2$ [using (1)]
$\Rightarrow2\vec{\text{a}}.\vec{\text{b}}=15$
$\Rightarrow2|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta=15$
$\Rightarrow2(3)(5)\cos\theta=15$ [using (1)]
$\Rightarrow\cos\theta=\frac{1}{2}$
$\therefore\theta=\frac{\pi}{3}$
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