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

Given the relation $$R$$ on $$\mathbb{N}$$ defined by $$a R b$$ if $$2a + 3b$$ is a multiple of $$5$$.

First, we show that $$R$$ is reflexive by checking whether $$a R a$$ holds. We have $$2a + 3a = 5a$$, which is clearly divisible by $$5$$, so $$R$$ is reflexive.

Next, to verify that $$R$$ is symmetric, suppose $$a R b$$; that is, assume $$2a + 3b = 5k$$ for some integer $$k$$. Consider the expression $$2b + 3a$$. We can write

$$2b + 3a = 5(a + b) - (2a + 3b) = 5(a + b) - 5k = 5(a + b - k).$$

Since this is a multiple of $$5$$, it follows that $$b R a$$, and hence $$R$$ is symmetric.

Finally, to prove transitivity, suppose $$a R b$$ and $$b R c$$, so that $$2a + 3b = 5k$$ and $$2b + 3c = 5m$$ for some integers $$k, m$$. Adding these two equations yields

$$2a + 5b + 3c = 5(k + m).$$

Rewriting, we have

$$2a + 3c = 5(k + m) - 5b = 5(k + m - b),$$

which is a multiple of $$5$$. Therefore $$a R c$$, and $$R$$ is transitive.

Since $$R$$ is reflexive, symmetric, and transitive, it is an equivalence relation. The correct answer is Option D.