Explanation
For any element $x_i$ present in $X$, 4 cases arise while making subsets $Y$ and $Z$.
Case 1: $x_i \in Y, x_i \in Z \implies Y \cap Z \neq \emptyset$
Case 2: $x_i \in Y, x_i \notin Z \implies Y \cap Z = \emptyset$
Case 3: $x_i \notin Y, x_i \in Z \implies Y \cap Z = \emptyset$
Case 4: $x_i \notin Y, x_i \notin Z \implies Y \cap Z = \emptyset$
Therefore, for every element, the number of ways is $3$ for which $Y \cap Z = \emptyset$.
Thus, the total number of ways is $3 \times 3 \times 3 \times 3 \times 3 = 3^5$ since the number of elements in set $X$ is $5$.