(d)
Since the given points lie in a plane.
$\therefore\ \left| \begin{matrix}
a & a & c \\
1 & 0 & 1 \\
c & c & b
\end{matrix} \right| = 0$
$Applying\ \ C_{1} \rightarrow C_{1} - C_{2}$
$\Longrightarrow \ \left| \begin{matrix}
0 & a & c \\
1 & 0 & 1 \\
0 & c & b
\end{matrix} \right| = 0$
$\Longrightarrow - 1\left( ab - c^{2} \right) = 0$
⟹ $c^{2} = ab$
Hence, $c$ is GM of $a\ and\ b$.