If $P(A \cup B \cup C) = 1$, $P(B) = 2P(A) $, $P(C) = 3P(A) $, $P(A \cap B) = P(A \cap C) = P(B \cap C) $, then $P(A) \le \frac14$

Let $\mathbb P(A)=a$, then $\mathbb P(B)=2a$ and $\mathbb P(C)=3a$. Hence $\mathbb P(B\cap C)\geqslant \mathbb P(B)+\mathbb P(C)-1$, that is, $\mathbb P(B\cap C)\geqslant5a-1$. By hypothesis, $\mathbb P(A\cap B)=\mathbb P(B\cap C)$ hence $\mathbb P(A\cap B)\geqslant5a-1$. But $\mathbb P(A\cap B)\leqslant \mathbb P(A)=a$ hence $a\geqslant5a-1$, that is, $\mathbb P(A)=a\leqslant\frac14$.

Note: This does not use the hypotheses on $\mathbb P(A\cup B\cup C)$ and $\mathbb P(A\cap C)$.