If $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$ and $\left|\overrightarrow{a}\right|=3,\left|\overrightarrow{b}\right|=5,\left|\overrightarrow{c}\right|=7$, then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is

The correct option is:
$\frac{\pi }{3}$

Step-by-step Derivation:

We are given the vector equation:
$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$
We are also given the magnitudes of the vectors:
$\left|\overrightarrow{a}\right|=3,\left|\overrightarrow{b}\right|=5,\left|\overrightarrow{c}\right|=7$

To find the angle between vectors $\overrightarrow{a}$ and $\overrightarrow{b}$, we can rearrange the given vector equation to isolate the vector $\overrightarrow{c}$ on one side:
$\overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}$

Next, we take the dot product of each side with itself (which is equivalent to squaring the magnitude of both sides):
${|\overrightarrow{a}+\overrightarrow{b}|}^2={|-\overrightarrow{c}|}^2$
Using the properties of the vector dot product, we can expand the left-hand side:
${\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2(\overrightarrow{a}·\overrightarrow{b})={\left|\overrightarrow{c}\right|}^2$

Recall that the dot product of two vectors is defined as:
$\overrightarrow{a}·\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta$
where $\theta$ is the angle between vectors $\overrightarrow{a}$ and $\overrightarrow{b}$. Substituting this definition into our expanded equation, we get:
${\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta ={\left|\overrightarrow{c}\right|}^2$

Now, we substitute the given numerical magnitudes into the equation:
$3^2+5^2+2\left(3\right)\left(5\right)\cos \theta =7^2$

Simplify each term:
$9+25+30\cos \theta =49$
$34+30\cos \theta =49$

Subtract 34 from both sides to isolate the term containing $\cos \theta$:
$30\cos \theta =49-34$
$30\cos \theta =15$

Divide by 30:
$\cos \theta =\frac{15}{30}=\frac{1}{2}$

Since the angle $\theta$ between two vectors lies in the interval $[0,\pi ]$, we find the principal value:
$\theta =\arccos \left(\frac{1}{2}\right)=\frac{\pi }{3}$
Therefore, the angle between the vectors is $\frac{\pi }{3}$ radians (or 60°).