Conditional Probability P(A intersect B intersect C)

It’s just a double application of the two-event formula, first thinking of $A\cap B$ as a single event:

$$\begin{align*} P(A\cap B\cap C)&=P\big((A\cap B)\cap C\big)\\ &=P\big(C\mid(A\cap B)\big)P(A\cap B)\\ &=P\big(C\mid(A\cap B)\big)\Big(P(B\mid A)P(A)\Big)\\ &=P(A)P(B\mid A)P(C\mid A\cap B)\;. \end{align*}$$

It is the same thing but you use the same rule two times.

$$P(A,B,C)=P(C\mid B,A)P(B,A)$$

Here assume $(A,B)=K$ then $P(A,B,C)=P(C\mid K)P(K)$ same with your rule. Then for the second case

$$P(A,B,C)=P(C\mid B,A)P(B,A)=P(C\mid B,A)P(B\mid A)P(A)$$

using again the same rule $P(B,A)=P(B\mid A)P(A)$.