For precipitating $\mathrm{Fe}(\mathrm{OH})_3$ sol, the best precipitating agent is

$\mathrm{Fe}(\mathrm{OH})_3$ is a positive sol, hence the best coagulant for precipitation of $\mathrm{Fe}(\mathrm{OH})_3$ is $\mathrm{Na}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$. The precipitating effect of an ion on dispersed phase of opposite charge increases with the valence of ions. According to Hardy-Schulze rule, greater is the valency of ions, greater is its power to bring about coagulation.
$$ \begin{aligned} \mathrm{Na}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right] & \longrightarrow 3 \mathrm{Na}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-} \\ \mathrm{Na}_2 \mathrm{SO}_4 & \longrightarrow 2 \mathrm{Na}^{+}+\mathrm{SO}_4^{2-} \\ \mathrm{Cr}(\mathrm{OH})_3 & \longrightarrow \mathrm{Cr}^{3+}+3 \mathrm{OH}^{-} \\ \mathrm{Al}(\mathrm{OH})_3 & \longrightarrow \mathrm{Al}^{3+}+3 \mathrm{OH}^{-} \end{aligned} $$ The ion with highest negative valency will coagulate positive sol of $\mathrm{Fe}(\mathrm{OH})_3$ is $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$. As it has highest coagulating power from relation (i), it would have lowest flocculating value and hence, is the best agent for coagulation.