Let R be a relation defined on $\mathrm{\mathbb{N}}$ as aRb if $2\mathrm{a}+3\mathrm{b}$ is a multiple of 5,  $\mathrm{a},\mathrm{b}\in \mathrm{\mathbb{N}}$. Then R is

$\mathrm{Let} (\mathrm{a},\mathrm{b})\in \mathrm{R}$ 
$\mathrm{f}(\mathrm{a},\mathrm{b})=2\mathrm{a}+3\mathrm{b}$
For Reflexive
$\begin{array}{l}\mathrm{f}(\mathrm{a},\mathrm{a})=2\mathrm{a}+3\mathrm{a}=5\mathrm{a}\text{ i,e divisible by }5\\ \Rightarrow (\mathrm{a},\mathrm{a})\in \mathrm{R}\end{array}$
For symmetric
$\mathrm{f}(\mathrm{b},\mathrm{a})=2\mathrm{b}+3\mathrm{a}=5\mathrm{a}+5\mathrm{b}-(2\mathrm{a}+3\mathrm{b}),divisible by 5$
    $$$\mathrm{f}(\mathrm{b},\mathrm{a})$is divisible by 5 $\Rightarrow (\mathrm{b},\mathrm{a})\in \mathrm{R}$
For transitive
$\mathrm{f}(\mathrm{a},\mathrm{b})=2\mathrm{a}+3\mathrm{b}$is divisible by 5
$\Rightarrow 2a+3b=5\alpha$
$\mathrm{f}(\mathrm{b},\mathrm{c})=2\mathrm{b}+3\mathrm{c}$ is divisible by 5
$\Rightarrow 2b+3c=5\beta$
$$\begin{gathered} 2a+5b+3c=5(\alpha +\beta ) \\ \begin{array}{r}\begin{array}{r}\Rightarrow 2a+3c=5(\alpha +\beta -b) \\ \Rightarrow \text{ a R c }\\ \end{array}\end{array} \end{gathered}$$
So, $2\mathrm{a}+3\mathrm{c}$ is divisible by 5
$\Rightarrow (\mathrm{a},\mathrm{c})\in \mathrm{R}$