Explanation

(c)
Let $\overrightarrow{\alpha} = \lambda\ \overrightarrow{\mathbf{a}} + \mu\overrightarrow{\mathbf{b}} + t\overrightarrow{\mathbf{c}}$ …(i)
Now, $\overrightarrow{\mathbf{a}}\mathbf{\bullet}\overrightarrow{\mathbf{p}}\mathbf{=}\overrightarrow{\mathbf{b}}\mathbf{\bullet}\overrightarrow{\mathbf{q}}\mathbf{=}\overrightarrow{\mathbf{c}}\mathbf{\bullet}\overrightarrow{\mathbf{r}} = 1$
$\Rightarrow \ \ \overrightarrow{\alpha}\mathbf{\bullet}\overrightarrow{\mathbf{p}} = \lambda\ \left( \mathbf{\ }\overrightarrow{\mathbf{a}}\mathbf{\bullet}\overrightarrow{\mathbf{p}} \right) + 0 + 0$
$\Rightarrow \ \lambda = \overrightarrow{\alpha} \bullet \overrightarrow{\mathbf{p}}$
Similarly, $\mu = \overrightarrow{\alpha} \bullet \overrightarrow{\mathbf{q}}$
and $t = \overrightarrow{\alpha} \bullet \overrightarrow{\mathbf{r}}$
From Eq. (i), we get
$\overrightarrow{\alpha} = \left( \ \overrightarrow{\alpha} \bullet \overrightarrow{\mathbf{p}} \right)\overrightarrow{\mathbf{a}} + \left( \ \overrightarrow{\alpha} \bullet \overrightarrow{\mathbf{q}}\ \right)\overrightarrow{\mathbf{b}} + (\overrightarrow{\alpha} \bullet \overrightarrow{\mathbf{r}})\overrightarrow{\mathbf{c}}$