Explanation

(c)
Since, $4a^{2} + 9b^{2} + 16c^{2} = 2(3ab + 6bc + 4ca)$
$\Rightarrow 8a^{2} + 18b^{2} + 32c^{2} - 12ab - 24bc - 16ca = 0$
$\Rightarrow \left( 4a^{2} - 12ab + 9b^{2} \right) + \left( 9b^{2} - 24bc + 16c^{2} \right) + \left( 16c^{2} - 16ca + 4a^{2} \right) = 0$
$\Rightarrow (2a - 3b)^{2} + (3b - 4c)^{2} + (4c - 2a)^{2} = 0$
$\Rightarrow 2a = 3b,\ 3b = 4c,\ 4c = 2a$
$\Rightarrow 2a = 3b = 4c = k$ [say]
$\Rightarrow a = \frac{k}{2},\ b = \frac{k}{3},\ c = \frac{k}{4}$
$\Rightarrow a,b,c$ are in HP.