Explanation

(c)
$\sin^{2}B - \sin^{2}A = \sin^{2}C - \sin^{2}B$ $\lbrack\because\ a^{2},b^{2},c^{2}$ are in AP]
$\Rightarrow \sin{(B + A)}\sin{(B - A)} = \sin{(C + B)}\sin{(C - B)}$
$\Rightarrow \sin C\left( \sin B\cos A - \cos B\sin A \right) = \sin A(\sin C\cos B - \cos C\sin B)$
$\Rightarrow 2\cot B = \cot A + \cot C$ [divide by $\sin A\sin B\sin C$]
$\Rightarrow \cot A,\cot B,\cot C$ are in AP.