Let $$R$$ be a relation on $$\mathbb{R}$$, given by $$R = \{a, b : 3a - 3b + \sqrt{7}$$ is an irrational number$$\}$$. Then $$R$$ is

Let $$R = \{(a, b) : 3a - 3b + \sqrt{7}$$ is an irrational number$$\}$$. We need to determine the properties of $$R$$.

To begin,

For $$(a, a)$$: $$3a - 3a + \sqrt{7} = \sqrt{7}$$, which is irrational. So $$(a, a) \in R$$ for all $$a \in \mathbb{R}$$. $$R$$ is reflexive.

Next,

If $$(a, b) \in R$$, then $$3a - 3b + \sqrt{7}$$ is irrational. For $$(b, a)$$: $$3b - 3a + \sqrt{7} = -(3a - 3b) + \sqrt{7}$$.

Take $$a = 0, b = \dfrac{\sqrt{7}}{3}$$: $$3(0) - 3\cdot\dfrac{\sqrt{7}}{3} + \sqrt{7} = -\sqrt{7} + \sqrt{7} = 0$$, which is rational. So $$(0, \dfrac{\sqrt{7}}{3}) \notin R$$.

Take $$a = \dfrac{\sqrt{7}}{3}, b = 0$$: $$3\cdot\dfrac{\sqrt{7}}{3} - 0 + \sqrt{7} = \sqrt{7} + \sqrt{7} = 2\sqrt{7}$$, which is irrational. So $$(\dfrac{\sqrt{7}}{3}, 0) \in R$$.

But $$(0, \dfrac{\sqrt{7}}{3}) \notin R$$. So $$R$$ is not symmetric.

From this,

Let $$a = \dfrac{\sqrt{7}}{3}, b = 0, c = \dfrac{-\sqrt{7}}{3}$$.

$$(a, b)$$: $$3 \cdot \dfrac{\sqrt{7}}{3} - 0 + \sqrt{7} = 2\sqrt{7}$$ (irrational) $$\checkmark$$

$$(b, c)$$: $$0 - 3\cdot\dfrac{-\sqrt{7}}{3} + \sqrt{7} = \sqrt{7} + \sqrt{7} = 2\sqrt{7}$$ (irrational) $$\checkmark$$

$$(a, c)$$: $$3\cdot\dfrac{\sqrt{7}}{3} - 3\cdot\dfrac{-\sqrt{7}}{3} + \sqrt{7} = \sqrt{7} + \sqrt{7} + \sqrt{7} = 3\sqrt{7}$$ (irrational) $$\checkmark$$

This particular case works, but let's try another: $$a = \dfrac{2\sqrt{7}}{3}, b = \dfrac{\sqrt{7}}{3}, c = 0$$.

$$(a, b)$$: $$2\sqrt{7} - \sqrt{7} + \sqrt{7} = 2\sqrt{7}$$ (irrational) $$\checkmark$$

$$(b, c)$$: $$\sqrt{7} - 0 + \sqrt{7} = 2\sqrt{7}$$ (irrational) $$\checkmark$$

$$(a, c)$$: $$2\sqrt{7} - 0 + \sqrt{7} = 3\sqrt{7}$$ (irrational) $$\checkmark$$

Let me find a counterexample. Take $$b = \dfrac{\sqrt{7}}{6}$$, $$a = 0$$, $$c = \dfrac{\sqrt{7}}{3}$$:

$$(a,b)$$: $$0 - \dfrac{\sqrt{7}}{2} + \sqrt{7} = \dfrac{\sqrt{7}}{2}$$ (irrational) $$\checkmark$$

$$(b,c)$$: $$\dfrac{\sqrt{7}}{2} - \sqrt{7} + \sqrt{7} = \dfrac{\sqrt{7}}{2}$$ (irrational) $$\checkmark$$

$$(a,c)$$: $$0 - \sqrt{7} + \sqrt{7} = 0$$ (rational) $$\times$$

So $$R$$ is not transitive.

The correct answer is Option A: Reflexive but neither symmetric nor transitive.