If $\overrightarrow{a} + \overrightarrow{b}$, $\overrightarrow{b} + \overrightarrow{c}$ and $\overrightarrow{c} + \overrightarrow{a}$ are coplanar, then show that the vectors $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ are also coplanar.
If $\overrightarrow{a} + \overrightarrow{b}$, $\overrightarrow{b} + \overrightarrow{c}$, $\overrightarrow{c} + \overrightarrow{a}$ are coplanar, then
$\left(\overrightarrow{a} + \overrightarrow{b}\right) \cdot \left[\left(\overrightarrow{b} + \overrightarrow{c}\right) \times \left(\overrightarrow{c} + \overrightarrow{a}\right)\right] = 0$
i.e. $\;$ $\left(\overrightarrow{a} + \overrightarrow{b}\right) \cdot \left[\left(\overrightarrow{b} \times \overrightarrow{c}\right) + \left(\overrightarrow{b} \times \overrightarrow{a}\right) + \left(\overrightarrow{c} \times \overrightarrow{c}\right) + \left(\overrightarrow{c} \times \overrightarrow{a}\right)\right] = 0$
Since $\overrightarrow{c} \times \overrightarrow{c} = 0$, we have
$\left(\overrightarrow{a} + \overrightarrow{b}\right) \cdot \left[\left(\overrightarrow{b} \times \overrightarrow{c}\right) + \left(\overrightarrow{b} \times \overrightarrow{a}\right) + \left(\overrightarrow{c} \times \overrightarrow{a}\right)\right] = 0$
i.e. $\;$ $\overrightarrow{a} \cdot \left(\overrightarrow{b} \times \overrightarrow{c}\right) + \overrightarrow{a} \cdot \left(\overrightarrow{b} \times \overrightarrow{a}\right) + \overrightarrow{a} \cdot \left(\overrightarrow{c} \times \overrightarrow{a}\right)$
$\hspace{2cm}$ $+ \overrightarrow{b} \cdot \left(\overrightarrow{b} \times \overrightarrow{c}\right) + \overrightarrow{b} \cdot \left(\overrightarrow{b} \times \overrightarrow{a}\right) + \overrightarrow{b} \cdot \left(\overrightarrow{c} \times \overrightarrow{a}\right) = 0$ $\;\;\; \cdots \; (1)$
Now,
$\overrightarrow{a} \cdot \left(\overrightarrow{b} \times \overrightarrow{a}\right) = - \overrightarrow{a} \cdot \left(\overrightarrow{a} \times \overrightarrow{b}\right) = - \left(\overrightarrow{a} \times \overrightarrow{a}\right) \cdot \overrightarrow{b} = 0$ $\;\;\; \cdots \; (2a)$
$\overrightarrow{a} \cdot \left(\overrightarrow{c} \times \overrightarrow{a}\right) = - \overrightarrow{a} \cdot \left(\overrightarrow{a} \times \overrightarrow{c}\right) = - \left(\overrightarrow{a} \times \overrightarrow{a}\right) \cdot \overrightarrow{c} = 0$ $\;\;\; \cdots \; (2b)$
$\overrightarrow{b} \cdot \left(\overrightarrow{b} \times \overrightarrow{c}\right) = \left(\overrightarrow{b} \times \overrightarrow{b}\right) \cdot \overrightarrow{c} = 0$ $\;\;\; \cdots \; (2c)$
$\overrightarrow{b} \cdot \left(\overrightarrow{b} \times \overrightarrow{a}\right) = \left(\overrightarrow{b} \times \overrightarrow{b}\right) \cdot \overrightarrow{a} = 0$ $\;\;\; \cdots \; (2d)$
$\overrightarrow{b} \cdot \left(\overrightarrow{c} \times \overrightarrow{a}\right) = \left(\overrightarrow{b} \times \overrightarrow{c}\right) \cdot \overrightarrow{a} = \overrightarrow{a} \cdot \left(\overrightarrow{b} \times \overrightarrow{c}\right)$ $\;\;\; \cdots \; (2e)$
$\therefore$ $\;$ In view of equations $(2a) ... (2e)$, equation $(1)$ becomes
$\overrightarrow{a} \cdot \left(\overrightarrow{b} \times \overrightarrow{c}\right) + \overrightarrow{a} \cdot \left(\overrightarrow{b} \times \overrightarrow{c} \right) = 0$
i.e. $\;$ $2 \overrightarrow{a} \cdot \left(\overrightarrow{b} \times \overrightarrow{c}\right) = 0$
i.e. $\;$ $\overrightarrow{a} \cdot \left(\overrightarrow{b} \times \overrightarrow{c}\right) = 0$
i.e. $\;$ $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ are coplanar.
$\therefore$ $\;$ If $\overrightarrow{a} + \overrightarrow{b}$, $\overrightarrow{b} + \overrightarrow{c}$, $\overrightarrow{c} + \overrightarrow{a}$ are coplanar, then $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ are also coplanar.