We are given the relation

$$R = \{(a,b)\in \mathbb R\times \mathbb R : 3a-3b+\sqrt{7}\text{ is irrational}\}.$$

We must determine whether this relation is reflexive, symmetric, and transitive.

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1. Simplify the condition

Observe that

$$3a-3b+\sqrt{7}=3(a-b)+\sqrt{7}.$$

Since $3(a-b)$ is a real number, the irrationality depends on whether adding it to $\sqrt7$ gives an irrational number.

A useful fact:

• irrational $+$ rational $=$ irrational
• irrational $+$ irrational may be rational or irrational

So we test the relation properties directly.

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2. Reflexive?

A relation is reflexive if $(a,a)\in R$ for every $a\in\mathbb R$.

Put $b=a$. Then

$$3a-3a+\sqrt7=\sqrt7,$$

which is irrational.

Hence $(a,a)\in R$ for every $a\in\mathbb R$.

Therefore, $R$ is reflexive.

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3. Symmetric?

A relation is symmetric if

$$(a,b)\in R \implies (b,a)\in R.$$

Suppose $(a,b)\in R$. Then

$$3a-3b+\sqrt7$$

is irrational.

Now check $(b,a)$:

$$3b-3a+\sqrt7 = -\bigl(3a-3b\bigr)+\sqrt7.$$

This need not be irrational whenever the first one is irrational.

We give a counterexample.

Take

$$a-b=-\frac{\sqrt7}{3}.$$

Then

$$3a-3b+\sqrt7 = 3\left(-\frac{\sqrt7}{3}\right)+\sqrt7=-\sqrt7+\sqrt7=0,$$

which is rational. So for symmetry we want $(a,b)\in R$ but $(b,a)\notin R$, so choose instead

$$a-b=\frac{\sqrt7}{3}.$$

Then

$$3a-3b+\sqrt7 = \sqrt7+\sqrt7 = 2\sqrt7,$$

which is irrational, so $(a,b)\in R$.

But then

$$3b-3a+\sqrt7 = -\sqrt7+\sqrt7=0,$$

which is rational, so $(b,a)\notin R$.

Hence, $R$ is not symmetric.

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4. Transitive?

A relation is transitive if

$$(a,b)\in R \text{ and } (b,c)\in R \implies (a,c)\in R.$$

We need a counterexample.

Choose

$$3a-3b=1-\sqrt7,$$

so that

$$3a-3b+\sqrt7 = 1,$$

which is rational. That would not help. So instead choose values making each pair irrational but the final one rational.

Let

$$3a-3b = 1, qquad 3b-3c = -1.$$

Then

$$3a-3b+\sqrt7 = 1+\sqrt7,$$

which is irrational, so $(a,b)\in R$.

Also,

$$3b-3c+\sqrt7 = -1+\sqrt7,$$

which is irrational, so $(b,c)\in R$.

But

$$3a-3c = (3a-3b)+(3b-3c)=1+(-1)=0,$$

so

$$3a-3c+\sqrt7 = \sqrt7,$$

which is irrational. This still gives transitivity in this example, so we need another one.

Let us construct it more carefully.

We want:

• $3a-3b+\sqrt7$ irrational,
• $3b-3c+\sqrt7$ irrational,
• $3a-3c+\sqrt7$ rational.

Take

$$3a-3b = 0, qquad 3b-3c = -\sqrt7.$$

Then

$$3a-3b+\sqrt7 = \sqrt7,$$

which is irrational.

But

$$3b-3c+\sqrt7 = -\sqrt7+\sqrt7 = 0,$$

which is rational, so that fails.

Try instead:

$$3a-3b = 1, qquad 3b-3c = -1-\sqrt7.$$

Then

$$3a-3b+\sqrt7 = 1+\sqrt7,$$

irrational.

Also

$$3b-3c+\sqrt7 = -1-\sqrt7+\sqrt7=-1,$$

which is rational, so not good.

A better approach is to rewrite the condition.

$(a,b)\in R$ iff

$$3(a-b)+\sqrt7 \text{ is irrational.}$$

Now if $3(a-b)$ is rational, then this is irrational. If $3(a-b)$ equals $r-\sqrt7$ for rational $r$, then the sum becomes rational.

So to violate transitivity, choose:

• $a-b$ rational/3,
• $b-c$ rational/3,
• but $a-c = (a-b)+(b-c)$ such that $3(a-c)+\sqrt7$ is rational.

This is impossible if both $3(a-b)$ and $3(b-c)$ are rational, because then $3(a-c)$ is rational and adding $\sqrt7$ remains irrational.

So choose one irrational difference cleverly.

Let

$$3a-3b = 1, qquad 3b-3c = -1-2\sqrt7.$$

Then

$$3a-3b+\sqrt7 = 1+\sqrt7,$$

irrational.

And

$$3b-3c+\sqrt7 = -1-2\sqrt7+\sqrt7=-1-\sqrt7,$$

irrational.

Now

$$3a-3c=(3a-3b)+(3b-3c)=1+(-1-2\sqrt7)=-2\sqrt7.$$

Hence

$$3a-3c+\sqrt7 = -2\sqrt7+\sqrt7=-\sqrt7,$$

which is irrational again. Still not a counterexample.

We need the sum to become of the form $r-\sqrt7$.

Take

$$3a-3b = 1, qquad 3b-3c = -1-\sqrt7.$$

Then

$$3a-3b+\sqrt7 = 1+\sqrt7,$$

irrational.

But

$$3b-3c+\sqrt7 = -1,$$

which is rational, so no.

Now use a direct simple construction:

Let

$$a=\frac{\sqrt7}{3},\quad b=0,\quad c=\frac{2\sqrt7}{3}.$$

Then

$$3a-3b+\sqrt7 = \sqrt7+\sqrt7=2\sqrt7,$$

irrational, so $(a,b)\in R$.

Also,

$$3b-3c+\sqrt7 = -2\sqrt7+\sqrt7=-\sqrt7,$$

irrational, so $(b,c)\in R$.

But

$$3a-3c+\sqrt7 = (\sqrt7-2\sqrt7)+\sqrt7=0,$$

which is rational. Hence $(a,c)\notin R$.

Therefore, $R$ is not transitive.

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5. Final classification

We found:

• Reflexive: Yes
• Symmetric: No
• Transitive: No

So $R$ is reflexive but neither symmetric nor transitive.

Thus the correct option is

$$\boxed{\text{D}}$$

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6. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.