Step 1: Recall the chain rule for probability.
For two events the multiplication law says $P(A\cap B)=P(A)\,P(B\mid A)$, where we first commit to $A$, then ask for $B$ given $A$.
Step 2: Extend the idea to three events.
Build the intersection one event at a time: first $A_1$, then $A_2$ given $A_1$, then $A_3$ given both. This yields $P(A_1\cap A_2\cap A_3)=P(A_1)\,P(A_2\mid A_1)\,P(A_3\mid A_1\cap A_2)$.
Step 3: Check the ordering of conditions.
Each new factor must be conditioned on everything already chosen, so the second factor conditions on $A_1$ and the third on $A_1\cap A_2$.
Step 4: Compare with the options.
We need a choice whose factors are $P(A_1)$, then a conditional of $A_2$ on $A_1$, then a conditional of $A_3$ on $A_1\cap A_2$.
Step 5: Reject the mismatched options.
The options that condition $A_1$ on $A_2$, or use only two factors, do not respect the build order and are therefore not always valid.
Step 6: Identify the correct expression.
Only $P(A_1)P(A_2/A_1)P(A_3/(A_1\cap A_2))$ matches the chain rule exactly.
\[ \boxed{P(A_1\cap A_2\cap A_3)=P(A_1)P(A_2/A_1)P(A_3/(A_1\cap A_2))} \]