Given: \(|A|=150,\ |B|=180,\ |C|=210,\ |D|=240\); every pair has 15 common elements; every triple has 3 common elements; and \(A\cap B\cap C\cap D=\varnothing\).

Use Inclusion–Exclusion for 4 sets:

\[ \begin{aligned} |A\cup B\cup C\cup D| &= \sum |A_i| \;-\; \sum |A_i\cap A_j| \;+\; \sum |A_i\cap A_j\cap A_k| \;-\; |A\cap B\cap C\cap D|\\ &= (150+180+210+240)\;-\; \binom{4}{2}\cdot 15 \;+\; \binom{4}{3}\cdot 3 \;-\; 0\\ &= 780 \;-\; 6\cdot 15 \;+\; 4\cdot 3\\ &= 780 - 90 + 12\\ &= \boxed{702}. \end{aligned} \]

Answer: 702