If $$x^2 + 9y^2 - 4x + 3 = 0$$, $$x, y \in R$$, then $$x$$ and $$y$$ respectively lie in the intervals

We begin with the given relation $$x^{2}+9y^{2}-4x+3=0$$ where both $$x$$ and $$y$$ are real numbers.

First, we bring all purely algebraic terms together and identify the quadratic expression in $$x$$ that can be completed to a perfect square. Re-writing, we have

$$x^{2}-4x+9y^{2}+3=0.$$

To complete the square in $$x$$, we recall the algebraic identity

$$(x-a)^{2}=x^{2}-2ax+a^{2}.$$

Comparing $$x^{2}-4x$$ with the right-hand side of the identity, we see that $$-2a=-4\; \Longrightarrow\; a=2.$$ Hence

$$x^{2}-4x=(x-2)^{2}-4.$$

Substituting this back into the original expression, we obtain

$$(x-2)^{2}-4+9y^{2}+3=0.$$

Now we combine the constant terms $$-4+3=-1$$ to get

$$(x-2)^{2}+9y^{2}-1=0.$$

Transposing $$-1$$ to the right side yields

$$(x-2)^{2}+9y^{2}=1.$$

This is the standard form of an ellipse centred at the point $$(2,\,0)$$. To make the domains of $$x$$ and $$y$$ explicit, we examine each variable in turn.

Because squares of real numbers are always non-negative, each term on the left must individually satisfy $$0 \le (x-2)^{2} \le 1$$ and $$0 \le 9y^{2} \le 1$$ so that their sum remains exactly $$1$$.

We start with the $$x$$-part:

$$0 \le (x-2)^{2} \le 1.$$

Taking square roots on the inequality $$0 \le (x-2)^{2} \le 1$$ gives

$$0 \le |x-2| \le 1.$$

This in turn implies

$$-1 \le x-2 \le 1.$$

Adding $$2$$ throughout, we obtain the interval for $$x$$:

$$1 \le x \le 3.$$

So $$x$$ necessarily lies in $$[1,\,3]$$.

Next we extract the allowable range of $$y$$ from the ellipse equation. Solving for $$y^{2}$$ we write

$$9y^{2}=1-(x-2)^{2}.$$

Dividing both sides by $$9$$ gives

$$y^{2}=\dfrac{1-(x-2)^{2}}{9}.$$

For real $$y$$, the right-hand side must be non-negative. But we have already ensured $$0 \le (x-2)^{2}\le 1,$$ so $$1-(x-2)^{2}\ge 0$$ automatically. Taking square roots yields

$$|y|=\sqrt{\dfrac{1-(x-2)^{2}}{9}} =\dfrac{\sqrt{1-(x-2)^{2}}}{3}\;.$$

The largest value of the numerator $$\sqrt{1-(x-2)^{2}}$$ is $$1$$, achieved when $$(x-2)^{2}=0$$, i.e. when $$x=2$$. Hence

$$|y|_{\text{max}}=\dfrac{1}{3}.$$

The smallest value is $$0$$, occurring when $$(x-2)^{2}=1$$, i.e. when $$x=1$$ or $$x=3$$. Combining these facts, we see that $$y$$ can range continuously between $$-\frac13$$ and $$+\frac13$$. Therefore

$$-\frac{1}{3}\le y\le \frac{1}{3}.$$

Summarising both intervals, we have

$$x\in [1,\,3]\quad\text{and}\quad y\in\left[-\dfrac13,\dfrac13\right].$$

Looking at the given choices, this matches exactly with Option B.

Hence, the correct answer is Option B.