Explanation

(c)
Given, $\overrightarrow{\alpha} + \overrightarrow{\beta} + \overrightarrow{\gamma} = a\overrightarrow{\delta}$ ….(i)
$\overrightarrow{\beta} + \overrightarrow{\gamma} + \overrightarrow{\delta} = b\overrightarrow{\alpha}$ …..(ii)
From Eq. (i)
$\overrightarrow{\alpha} + \overrightarrow{\beta} + \overrightarrow{\gamma}$ $+ \overrightarrow{\delta} = (a + 1)\overrightarrow{\delta}$ ...(iii)
From Eq. (ii)
$\overrightarrow{\alpha} + \overrightarrow{\beta} + \overrightarrow{\gamma}$ $+ \overrightarrow{\delta} = (b + 1)\overrightarrow{\alpha}$ …(iv)
From Eq. (iii) and (iv),
$(a + 1)\overrightarrow{\delta} = (b + 1)\overrightarrow{\alpha}$ …(v)
Since, $\overrightarrow{\alpha}$ is not parallel to $\overrightarrow{\delta}$.
$\therefore$ From Eq. (v),
$a + 1 = 0\ and\ b + 1 = 0$
$\therefore$ From Eq. (iii),
$\overrightarrow{\alpha} + \overrightarrow{\beta} + \overrightarrow{\gamma}$ $+ \overrightarrow{\delta} = \overrightarrow{0}$