Step 1: Use the precipitation data to split the formula.
Three moles of $AgCl$ forming per mole of complex tells us straightaway that all three chloride ions are sitting outside the coordination sphere, free to react with $AgNO_3$. That leaves all six $NH_3$ molecules bound directly to cobalt inside the sphere, giving the structural formula $[Co(NH_3)_6]Cl_3$.
Step 2: Work out cobalt's oxidation state and name the compound.
Ammonia is neutral and the three chlorides outside carry a total charge of $-3$, so for the whole compound to be neutral, cobalt must be $+3$. Naming it in order, six ammine ligands, cobalt in the $+3$ state, and chloride as the counter ion, gives hexaamminecobalt(III) chloride.
Step 3: Sort out the electron configuration and hybridisation.
Free $Co^{3+}$ has the configuration $[Ar]3d^6$. Ammonia is a strong field ligand, so it forces all six of these d-electrons to pair up within just three of the five d-orbitals, leaving two d-orbitals completely empty and available for bonding. Those two empty d-orbitals combine with one s and three p orbitals, giving $d^2sp^3$ hybridisation and an octahedral, inner-orbital complex.
Step 4: Read off the magnetic behaviour.
Since every one of the six d-electrons is now paired up, there are no unpaired electrons left anywhere on cobalt, so the complex shows no attraction to a magnetic field, it is diamagnetic.
\[ \boxed{\text{[Co(NH}_3\text{)}_6\text{]Cl}_3\text{, hexaamminecobalt(III) chloride, } d^2sp^3\text{, diamagnetic}} \]