(c) Point $P$ will lie near the charge which is smaller in magnitude $i.e. - 6\mu C$. Hence potential at $P$
$V = \frac{1}{4\pi\varepsilon_{0}}\frac{\left( - 6 \times 10^{- 6} \right)}{x} + \frac{1}{4\pi\varepsilon_{0}}\frac{(12 \times 10^{- 6})}{(0.2 + x)} = 0 \Rightarrow x = 0.2\ m$