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

We are given the vector equation $$\vec{a}+\vec{b}+\vec{c}=0$$ together with the magnitudes $$|\vec{a}|=3,\;|\vec{b}|=5,\;|\vec{c}|=7$$. Let $$\theta$$ be the angle between $$\vec{a}$$ and $$\vec{b}$$. We want $$\theta$$.

From $$\vec{a}+\vec{b}+\vec{c}=0$$ we can write $$\vec{c}=-(\vec{a}+\vec{b})$$. Taking magnitudes on both sides and using the formula for the magnitude of a sum of vectors,

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

Recall the dot-product in terms of the included angle: $$\vec{a}\cdot\vec{b}=|\vec{a}|\,|\vec{b}|\,\cos\theta$$. Substitute the given magnitudes and this dot-product into the previous equation:

$$7^{2}=3^{2}+5^{2}+2(3)(5)\cos\theta$$
$$49=9+25+30\cos\theta$$

Isolate $$\cos\theta$$:

$$49-34=30\cos\theta$$
$$15=30\cos\theta$$
$$\cos\theta=\frac{15}{30}=\frac{1}{2}$$

The angle whose cosine is $$\frac{1}{2}$$ is $$\theta=\frac{\pi}{3}$$.

Hence, the angle between $$\vec{a}$$ and $$\vec{b}$$ is $$\boxed{\dfrac{\pi}{3}}$$.

Option A which is: $$\frac{\pi}{3}$$