A point $$P$$ moves on the line $$2x - 3y + 4 = 0$$. If $$Q(1, 4)$$ and $$R(3, -2)$$ are fixed points, then the locus of the centroid of $$\triangle PQR$$ is a line:

Let the moving point be $$P(x,\,y)$$. According to the question, $$P$$ always satisfies the straight-line equation $$2x - 3y + 4 = 0$$.

We also have two fixed points: $$Q(1,\,4)$$ and $$R(3,\,-2)$$.

For any triangle with vertices $$P(x,\,y)$$, $$Q(1,\,4)$$ and $$R(3,\,-2)$$, the centroid formula states:

$$\bigl(X,\,Y\bigr)\;=\;\Bigl(\dfrac{x + 1 + 3}{3},\;\dfrac{y + 4 - 2}{3}\Bigr).$$

So the centroid $$G$$ of $$\triangle PQR$$ has coordinates

$$X \;=\;\dfrac{x + 4}{3}, \qquad Y \;=\;\dfrac{y + 2}{3}.$$

Next, we use the condition that $$P(x,\,y)$$ lies on the line $$2x - 3y + 4 = 0$$. Rearranging that equation, we obtain

$$2x - 3y + 4 = 0 \;\;\Longrightarrow\;\; 2x = 3y - 4 \;\;\Longrightarrow\;\; x = \dfrac{3y - 4}{2}.$$

We substitute this expression for $$x$$ into the formula for $$X$$:

$$X \;=\;\dfrac{\dfrac{3y - 4}{2} + 4}{3}.$$

Combining the terms in the numerator,

$$\dfrac{3y - 4}{2} + 4 \;=\;\dfrac{3y - 4 + 8}{2} \;=\;\dfrac{3y + 4}{2}.$$

Thus

$$X \;=\;\dfrac{\dfrac{3y + 4}{2}}{3} \;=\;\dfrac{3y + 4}{6}.$$

From the expression for $$Y$$ we have

$$Y \;=\;\dfrac{y + 2}{3} \;\;\Longrightarrow\;\; y \;=\;3Y - 2.$$

We now eliminate $$y$$ by replacing it in the equation for $$X$$:

$$X \;=\;\dfrac{3(3Y - 2) + 4}{6}.$$

Expanding the numerator gives

$$3(3Y - 2) + 4 \;=\;9Y - 6 + 4 \;=\;9Y - 2,$$

so

$$X \;=\;\dfrac{9Y - 2}{6}.$$

To obtain the relation between $$X$$ and $$Y$$, we cross-multiply:

$$6X \;=\;9Y - 2.$$

Re-arranging,

$$9Y \;=\;6X + 2 \;\;\Longrightarrow\;\; Y = \dfrac{6}{9}X + \dfrac{2}{9}.$$

Simplifying the coefficient,

$$Y \;=\;\dfrac{2}{3}X + \dfrac{2}{9}.$$

This is the equation of a straight line whose slope is $$\dfrac{2}{3}$$. Therefore, the locus of the centroid $$G$$ is a line with slope $$\frac{2}{3}$$.

Hence, the correct answer is Option A.