Your unbalanced equation is
$"Na"_3"PO"_4 + "CaCl"_2 → "Ca"_3("PO"_4)_2 + "NaCl"$
1. Start with the most complicated formula, $"Ca"_3("PO"_4)_2$.
Put a 1 in front of it. We have now fixed the number of $"Ca"$ atoms and $"PO"_4$ groups.
$"Na"_3"PO"_4 + "CaCl"_2 →color(red)(1)"Ca"_3("PO"_4)_2 + "NaCl"$
2. Balance $"Ca"$. Put a 3 in front of $"CaCl"_2$.
$"Na"_3"PO"_4 + color(blue)(3)"CaCl"_2 → color(red)(1)"Ca"_3("PO"_4)_2 + "NaCl"$
.
3. Balance $"PO"_4$. Put a 2 in front of $"Na"_3"PO"_4$.
$color(orange)(2)"Na"_3"PO"_4 + color(blue)(3)"CaCl"_2 → color(red)(1) "Ca"_3("PO"_4)_2 + "NaCl"$
4. Balance $"Na"$. Put a 6 in front of $"NaCl"$.
$color(orange)(2) "Na"_3"PO"_4 + color(blue)(3) "CaCl"_2 → color(red)(1)"Ca"_3("PO"_4)_2 + color(teal)(6) "NaCl"$
5. Every formula has a coefficient.
The equation should be balanced. Let's check.
Left hand side: $"6 Na; 2 PO"_4"; 3 Ca; 6 Cl"$
Right hand side: $"3 Ca; 2 PO"_4"; 6 Na; 6 Cl"$.
Everything checks. the balanced equation is
$"2Na"_3"PO"_4 + "3CaCl"_2 → "Ca"_3("PO"_4)_2 + "6NaCl"$