Solution:

$\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{A}$ and $|\mathrm{a}-\mathrm{b}|$ is even $\}$ where $\mathrm{A}=$
(As given)
If $\mathrm{R}$ is Reflexive, Symmetric and Transitive, therefore $\mathrm{R}$ equivalence relation.

Reflexivity:
Suppose $a$ be an arbitrary element of A
$|\mathrm{a}-\mathrm{a}|=0$
As 0 is even
$\Rightarrow(\mathrm{a}, \mathrm{a}) \in \mathrm{R}$
Therefore, $\mathrm{R}$ is reflexive.

Symmetric:
Suppose $a$ and $b \in A$, such that $(a, b) \in R$
$\begin{array}{l}
\Rightarrow|\mathrm{a}-\mathrm{b}|=\text { even } \\
\Rightarrow|\mathrm{b}-\mathrm{a}|=\mathrm{even} \\
\Rightarrow(\mathrm{b}, \mathrm{a}) \in \mathrm{R}
\end{array}$
Therefore, $R$ is symmetric.

Transitivity:
Suppose $a, b$ and $c \in A$, such that $(a, b) \in R$ and $(b, c) \in R$
$\Rightarrow|\mathrm{a}-\mathrm{b}|$ is even and $|\mathrm{b}-\mathrm{c}|$ is even
This is only possible when $a$ and $b$ both are even or odd and $\mathrm{b}$ and $\mathrm{c}$ both are even or odd.
Two cases arise:

Case I: If $b$ is an even
Suppose $(a, b) \in R$ and $(b, c) \in R$
$\Rightarrow|a-b|$ is even and $|b-c|$ is even
As $b$ is even
$\Rightarrow$ a is even and $c$ is even
$\Rightarrow|\mathrm{a}-\mathrm{c}|$ is even
$\Rightarrow(a, c) \in R$

Case II: If b is an odd
Suppose $(\mathrm{a}, \mathrm{b}) \in \mathrm{R}$ and $(\mathrm{b}, \mathrm{c}) \in \mathrm{R}$
$\Rightarrow|\mathrm{a}-\mathrm{b}|$ is even and $|\mathrm{b}-\mathrm{c}|$ is even
$\Rightarrow$ As $b$ is odd
$\Rightarrow \mathrm{a}$ is odd and $\mathrm{c}$ is odd
As difference of two odd numbers is even
$\Rightarrow|\mathrm{a}-\mathrm{c}|$ is even
$\Rightarrow(\mathrm{a}, \mathrm{c}) \in \mathrm{R}$
Therefore, R is transitive.
As a result, R is an equivalence relation.