A = {1, 2, 3, …10}, S be the set of subset of A and R = {(a, b) : a, b ∈ S and a ∩ b ≠ ϕ}, then R is

The correct option is Symmetric only.

Analysis of the Given Data:
We are given:
• A set $A=\{1,2,3,\dots ,10\}$
• $S$ is the set of all subsets of $A$ (the power set of $A$). Note that the empty set $\varnothing$ is a subset of any set, so $\varnothing \in S$.
• A relation $R$ defined on $S$ such that:

$$R=\left\{\right(a,b):a,b\in S\text{ and }a\cap b\ne \varnothing \}$$

This means two subsets $a$ and $b$ are related under $R$ if and only if they share at least one common element (i.e., their intersection is non-empty).

Let us analyze the properties of the relation $R$ step-by-step:

1. Reflexivity:
A relation $R$ on a set $S$ is reflexive if $(a,a)\in R$ for all $a\in S$.
Let us check if this holds for the empty set $\varnothing$.
Since $\varnothing \in S$, we compute its intersection with itself:

$$\varnothing \cap \varnothing =\varnothing$$

Since the intersection is empty, it does not satisfy the condition $a\cap a\ne \varnothing$.
Thus, $(\varnothing ,\varnothing )\notin R$.
Since there exists at least one element in $S$ that is not related to itself, the relation $R$ is not reflexive.

2. Symmetry:
A relation $R$ is symmetric if $(a,b)\in R$ implies $(b,a)\in R$ for all $a,b\in S$.
Assume $(a,b)\in R$. By definition, this means:

$$a\cap b\ne \varnothing$$

Since set intersection is commutative, we know that:

$$b\cap a=a\cap b$$

Therefore, we have:

$$b\cap a\ne \varnothing$$

This implies $(b,a)\in R$.
Hence, the relation $R$ is symmetric.

3. Transitivity:
A relation $R$ is transitive if $(b,c)\in R$ and $(c,d)\in R$ implies $(b,d)\in R$ for all $b,c,d\in S$.
Let us choose three specific subsets as a counterexample:
• $b=\{1,2\}$
• $c=\{1,3,4\}$
• $d=\{4,5,6\}$
All these sets are subsets of $A$, so they belong to $S$.
Now let's check their intersections:
• For $b$ and $c$:

$$b\cap c=\{1,2\}\cap \{1,3,4\}=\left\{1\right\}\ne \varnothing$$

Since $b\cap c\ne \varnothing$, we have $(b,c)\in R$.
• For $c$ and $d$:

$$c\cap d=\{1,3,4\}\cap \{4,5,6\}=\left\{4\right\}\ne \varnothing$$

Since $c\cap d\ne \varnothing$, we have $(c,d)\in R$.
• For $b$ and $d$:

$$b\cap d=\{1,2\}\cap \{4,5,6\}=\varnothing$$

Since $b\cap d=\varnothing$, we have $(b,d)\notin R$.
Since $(b,c)\in R$ and $(c,d)\in R$ does not imply that $(b,d)\in R$, the relation $R$ is not transitive.

Conclusion:
Because the relation $R$ is symmetric but not reflexive and not transitive, it is classified as Symmetric only.