If $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ are three unit vectors such that $\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = \overrightarrow{0}$ find the value of $\overrightarrow{a} . \overrightarrow{b}$ + $\overrightarrow{b} . \overrightarrow{c}$ + $\overrightarrow{c} . \overrightarrow{a}$

The correct answer is $-\dfrac{3}{2}$

$|\overrightarrow{a}|$ = $|\overrightarrow{b}|$ = $|\overrightarrow{c}|$ = 1

∴ $(\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c})^2 = 0$

⇒ $|\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + |\overrightarrow{c}|^2$ + $2 \overrightarrow{a} . \overrightarrow{b}$ + $2 \overrightarrow{b} . \overrightarrow{c}$ + $2 \overrightarrow{c} . \overrightarrow{a}$ = $0$

∴ $1^2 + 1^2 + 1^2$ + $2 [\overrightarrow{a} . \overrightarrow{b} + \overrightarrow{b} . \overrightarrow{c} + \overrightarrow{c} . \overrightarrow{a}]$ = $0$
⇒ $[\overrightarrow{a} . \overrightarrow{b} + \overrightarrow{b} . \overrightarrow{c} + \overrightarrow{c} . \overrightarrow{a}]$ = $-\dfrac{3}{2}$