Note that $$A = (A\cap B^c \cap C^c) \bigcup (A \cap B \cap C^c) \bigcup (A\cap B^c \cap C) \bigcup (A \cap B \cap C)$$ In the above, we have written $A$ as the union of four disjoint sets. Hence, we get that $$\#A = \#(A\cap B^c \cap C^c) + \#(A \cap B \cap C^c) + \#(A\cap B^c \cap C) + \#(A \cap B \cap C) \,\,\,\,\,\,\,\, (\star)$$ Now note that we can write $A \cap B$ and $A \cap C$ as a disjoint union as follows. $$(A \cap B) = (A \cap B \cap C^c) \bigcup (A \cap B \cap C)$$ and $$(A \cap C) = (A \cap B \cap C) \bigcup (A \cap B^c \cap C)$$ Hence, we get that $$\#(A \cap B) = \#(A \cap B \cap C^c) + \#(A \cap B \cap C)$$ and $$\#(A \cap C) = \#(A \cap B \cap C) + \#(A \cap B^c \cap C)$$ Rearranging, we get that $$\#(A \cap B \cap C^c) = \#(A \cap B) - \#(A \cap B \cap C)$$ $$\#(A \cap B^c \cap C) = \#(A \cap C) - \#(A \cap B \cap C)$$ Plugging the above two in $(\star)$, we get that $$\#A = \#(A\cap B^c \cap C^c) + \#(A \cap B) + \#(A \cap C) - \#(A \cap B \cap C) \,\,\,\,\,\,\, (\dagger)$$ Plugging in the given values in $(\dagger)$, we get $$\#A = \#(A\cap B^c \cap C^c) + 11 + 12 - 5 = \#(A\cap B^c \cap C^c) + 18$$ Note that $\#X \geq 0$ for any set $X$ and hence, the minimum value of $\#A$ is when $\#(A\cap B^c \cap C^c) = 0$, which gives us $$\#A = 18$$ Note that the minimum is attained when $A \subseteq B \cup C$.