$R = \left\{ {\left( {x,y} \right):x,y \in N\,\,and\,\,{x^2} – 4xy + 3{y^2} = 0} \right\}$

Now, ${{x^2} – 4xy + 3{y^2} = 0}$

$ \Rightarrow \left( {x – y} \right)\left( {x – 3y} \right) = 0$

$\therefore x = y$ or $x = 3y$

$\therefore R = \left\{ {\left( {1,1} \right),\left( {3,1} \right),\left( {2,2} \right),\left( {6,2} \right),\left( {3,3} \right),\left( {9,3} \right),…..} \right\}$

Since $\left( {1,1} \right),\left( {2,2,} \right),\left( {3,3} \right)…..$ are preset in the relation, therefore $R$ is reflxive.

Since $(3,1)$ is an element of $R$ but $(1,3)$ is not the element of $R$, therefore $R$ is not symmetric

Here $\left( {3,1} \right) \in R$ and $\left( {1,1} \right) \in R \Rightarrow \left( {3,1} \right) \in R$

$\left( {6,2} \right) \in R$ and $\left( {2,2} \right) \in R \Rightarrow \left( {6,2} \right) \in R$

For all such $\left( {a,b} \right) \in R$ and $\left( {b,c} \right) \in R \Rightarrow \left( {a,b} \right) \in R$

Hence $R$ is transitive.