We have, $\overline{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overline{\mathrm{c}}=0$ ...(i)
On squaring both sides request.
$\bar{a}^{2}+\vec{b}^{2}+\bar{c}^{2}+2 \bar{a} \vec{b}+2 \vec{b} \bar{c}+2 \vec{c} \bar{a}=0$
$\quad(\because \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}, \vec{b} \cdot \vec{c}=\bar{c} \vec{b}$ and $\bar{c} \cdot \vec{a}=\bar{a} \cdot \vec{c})$
$\Rightarrow|a|^{2}+|b|^{2}+|c|^{2}=-2[a \cdot b+b . c+c . a]$
$\Rightarrow(3)^{2}+(5)^{2}+(7)^{2}=-2[a \cdot b+b \cdot c+c . a]$
$\Rightarrow a \cdot b+b . c+c . a=\frac{9+25+49}{-2}=-\frac{83}{2}$
Now $a+b+c=0 \quad$ [using eq. (i)]
$\Rightarrow \mathrm{a}+\mathrm{b}=-\mathrm{c}$
On squaring both sides, we get $\Rightarrow \mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{a} \cdot \mathrm{b}=\mathrm{c}^{2}$
$\Rightarrow(3)^{2}+(5)^{2}+2 \mathrm{ab}=(7)^{2}$
$\Rightarrow \overline{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\frac{15}{2}$
$\Rightarrow|\overline{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta=\frac{15}{2} \Rightarrow 3.5 \cos \theta=\frac{15}{2}$
$\Rightarrow \cos \theta=\frac{1}{2}=\cos \frac{\pi}{3}$
$\theta=\frac{\pi}{3}$
From eq. (i), $\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=-\overline{\mathrm{a}}$
$\Rightarrow b^{2}+c^{2}+2 b \cdot c=a^{2}$
$\Rightarrow 2 \mathrm{~b} \cdot \mathrm{c}=\mathrm{a}^{2}-\mathrm{b}^{2}-\mathrm{c}^{2}=9-25-49=-65$
$\Rightarrow \mathrm{b.c}=-\frac{65}{2} \Rightarrow|\mathrm{b}||\mathrm{c}| \cos \theta=-\frac{65}{2}$
$\Rightarrow \cos \theta=\frac{65}{2} \times \frac{1}{5} \times \frac{1}{7}=-\frac{13}{14}$
Also, $|\vec{a}+\vec{b}|=|-\vec{c}|=|\vec{c}|=7$