The easiest way to balance this chemical equation is to go by ions, not by atoms.
If you write the complete ionic equation for this reaction, you'll get
3NH_(4(aq))^(+) + PO_(4(aq))^(3-) + Ca_((aq))^(2+) + 2Cl_((aq))^(-) -> Ca_3(PO_4)_(2(s)) + NH_text(4(aq])^(+) + Cl_((aq))^(-)3NH+4(aq)+PO3−4(aq)+Ca2+(aq)+2Cl−(aq)→Ca3(PO4)2(s)+NH+4(aq]+Cl−(aq)
If you were to eliminate spectator ions, i.e. the ions that are present on both sides of the equation, you'll get the net ionic equation, which looks like this
PO_(4(aq))^(3-) + Ca_((aq))^(2+) -> Ca_3(PO_4)_(2(s))PO3−4(aq)+Ca2+(aq)→Ca3(PO4)2(s)
Notice that you need 3 calcium atoms on the products' side, but only have 1 calcium cation of the reactants' side ->→ multiply the calcium cation by 3.
Likewise, multiply the phosphate anions by 2 to get them to match the number of phosphate ions present on the products' side.
This will get you
color(red)(2)PO_(4(aq))^(3-) + color(blue)(3)Ca_((aq))^(2+) -> Ca_3(PO_4)_(2(s))2PO3−4(aq)+3Ca2+(aq)→Ca3(PO4)2(s)
Now take these stoichiometric coefficients and use them in the overall equation
color(red)(2)(NH_4)_3PO_text(4(aq]) + color(blue)(3)CaCl_(2(aq)) -> Ca_3(PO_4)_(2(s)) + NH_4Cl_((aq))2(NH4)3PO4(aq]+3CaCl2(aq)→Ca3(PO4)2(s)+NH4Cl(aq)
Now you only have to balance the ammonium ions, NH_4^(+)NH+4, and the ClCl. Notice that you have 6 of each on the reactants' side, so multiply the ammonium chloride by 6 to balance everything out
color(red)(2)(NH_4)_3PO_text(4(aq]) + color(blue)(3)CaCl_(2(aq)) -> Ca_3(PO_4)_(2(s)) + color(green)(6)NH_4Cl_((aq))2(NH4)3PO4(aq]+3CaCl2(aq)→Ca3(PO4)2(s)+6NH4Cl(aq)
