Let $A = \{a,b,c\},(R\;\&\;S) \subseteq A \times A$
$\implies (R\;\&\;S) \subseteq \{(a,a),(b,b),(c,c),(a,b)(b,a),(a,c),(c,a),(b,c),(c,b)\}$
A relation $R$ on a set $A$ is said to be a symmetric relation iff $(a,b)\in R\implies (b,a) \in R$ for all $a,b\in A$
$i.e., aRb\implies bRa$ for all $a,b \in A.$
Let, $(a,b) \in R,$ and $(a,b) \in S$
$(b,a) \in R,$ and $(b,a) \in S$
From above, we can conclude that, $(a,b)\in (R \cap S),$ and $(b,a) \in (R \cap S).$
$\therefore R\cap S$ is symmetric relations.
Again, let, $(a,b) \in R,$ or $(a,b) \in S$
$(b,a) \in R,$ or $(b,a) \in S$
From above, we can conclude that, $(a,b)\in (R \cup S),$ or $(b,a) \in (R \cup S).$
$\therefore R\cup S$ is symmetric relations.
So, the correct answer is $(A).$
