Each of the angle between vectors $\overrightarrow{a},\overrightarrow{b}$and$\overrightarrow{c}$, is equal to ${60}^{\circ }.$if $\left|\overrightarrow{a}\right|=4,\left|\overrightarrow{b}\right|=2$ and $\left|\overrightarrow{c}\right|=6$then the modulus of $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$, is

We know that 

$|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{|}^2$

$=|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2+2(\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{c}\cdot \overrightarrow{a}) .....(1)$

we find that 

$\begin{array}{l}\overrightarrow{a}\cdot \overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \frac{\pi }{3}=4\times 2\times \frac{1}{2}=4\\ \overrightarrow{b}\cdot \overrightarrow{c}=\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|\cos \frac{\pi }{3}=2\times 6\times \frac{1}{2}=6\\ \overrightarrow{c}\cdot \overrightarrow{a}=\left|\overrightarrow{c}\right|\left|\overrightarrow{a}\right|\cos \frac{\pi }{3}=6\times 4\times \frac{1}{2}=12\end{array}$

Substituting these values and $\left|\overrightarrow{a}\right|=4,\left|\overrightarrow{b}\right|=2\text{ and }\left|\overrightarrow{c}\right|=6$ in (I) we obtian.

$\begin{array}{l}|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{|}^2=4^2+2^2+6^2+2(4+6+12)=100\\ \Rightarrow |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=10\end{array}$