Reflexivity For reflexive $(x, x) \in R .$
$x+2 x=3 x$
which is divisible by 3.
$\Rightarrow$ xRx $\in R$ Hence, xRy is reflexive.
Symmetric Let $x+2 y=3 \lambda$
$\Rightarrow \quad x=3 \lambda-2 y$
$\begin{aligned}
\text {Now, } & \begin{aligned}
y+2 x &=y+2(3 \lambda-2 y) \\
&=y+6 \lambda-4 y \\
&=6 \lambda-3 y
\end{aligned}
\end{aligned}$
$\Rightarrow \quad y+2 x=3(2 \lambda-y)$
which is divisible by 3.
$\therefore$ $(x, y) \in R \Rightarrow(y, x) \in R$
Hence, xRy is symmetric.
$\begin{array}{ll}\text {Transifive } & x+2 y=3 \lambda \\ \text { and } & y+2 z=3 \mu\end{array}$
$x+3 y+2 z=3 x+3 \mu$
$\Rightarrow \quad x+2 z=3(\lambda+\mu-y)$
which is divisible by $3 .$
$\because (x, y) \in$ R and $(y, z) \in R$
$\Rightarrow \quad(x, z) \in R$
Hence, xRy is transitive
$\because$ R is reflexive, symmetric and transitive.
$\therefore$ R is an equivalence relation.