Consider $E = B \cup C$ so that

$P ( A \cup B \cup C ) = P ( A \cup E )$

$= P ( A )+ P ( E )- P ( A \cap E )$                …… $(1)$

Now

$P ( E )= P ( B \cup C )$

$= P ( B )+ P ( C )- P ( B \cap C )$               ……… $(2)$

Also $A \cap E=A \cap(B \cup C)$ $=(A \cap B) \cup(A \cap C)$   [using distribution property of intersection of sets over the union]. Thus

$P(A \cap E)=P(A \cap B)+P(A \cap C)$ $-P[(A \cap B) \cap(A \cap C)]$

$= P ( A \cap B )+ P ( A \cap C )- P [ A \cap B \cap C ] $        ……… $(3)$

Using $(2)$ and $( 3 )$ in $(1)$, we get

$P [ A \cup B \cup C ]= P ( A )+ P ( B )$ $+ P ( C )- P ( B \cap C )$ $- P ( A \cap B )- P ( A \cap C )$ $+ P ( A \cap B \cap C )$