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 $A \subseteq B \cup C$.