Let n(X) denote the number of elements in X.
Then, n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B) −n(B∩C)−n(C∩A)+n(A∩B∩C) =∑n(A)−∑n(A∩B)(∵A∩B∩C=ϕ)
Now, AΔB=(A−B)∪(B−A)=(A∪B)−(A∩B)
Therefore n(AΔB)=n(A∪B)−n(A∩B)=n(A)+n(B)−2n(A∩B)
and 300=∑n(AΔB)=∑[n(A)+n(B)−2n(A∩B)] =2[∑n(A)−∑n(A∩B)]
Therefore n(A∪B∪C)=∑n(A)−∑n(A∩B)=300/2=150