Which of the following is/are correct for potassium ferrocyanide?

Potassium ferrocyanide,

${\mathrm{K}}_4{\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}_3$ (Prussian blue)

It gives a brown precipitate with ${\mathrm{Cu}}^{2+}$ ions.

${\mathrm{K}}_4{\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}_3$$+{\mathrm{Cu}}^{2+}⟶{\mathrm{Cu}}_2\underset{\mathrm{Raddish}-\mathrm{Brown} \mathrm{ppt}}{\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}\downarrow$

It gives a white precipitate 9f mixed salt with ${\mathrm{Ca}}^{2+}$ ions.

${\left[\mathrm{FeC}(\mathrm{N}{)}_6\right]}^{4^{-}}+{\mathrm{Ca}}^{2+}\underset{{\mathrm{NH}}_4\mathrm{Cl}}{\to }\underset{\mathrm{White} \mathrm{ppt}}{\mathrm{Ca}\left({\mathrm{NH}}_4\right)\mathrm{K}}{\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}_6$

It in excess gives a bluish white/white precipitate with ${\mathrm{Zn}}^{2+}$

Thus, options  A, B and C are correct .