A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is:

Let us denote the required straight line by the equation $$L$$. Because this line meets the $$x$$-axis at the point $$P(a,0)$$ and the $$y$$-axis at the point $$Q(0,b)$$, its equation can be written in the intercept form

$$\frac{x}{a}+\frac{y}{b}=1.$$

We are told that the line always passes through the fixed point $$(2,3)$$. Substituting $$x=2,\;y=3$$ in the above equation gives

$$\frac{2}{a}+\frac{3}{b}=1.$$

Next, we complete the rectangle $$OPRQ$$ with $$O(0,0)$$ at the origin, $$P(a,0)$$ on the $$x$$-axis, and $$Q(0,b)$$ on the $$y$$-axis. In such an axis-parallel rectangle, the fourth (opposite) vertex $$R$$ has the coordinates $$R(a,b)$$ because the sides $$OP$$ and $$OQ$$ are respectively horizontal and vertical.

To describe the locus of $$R$$ we simply rename its coordinates:

$$x=a,\qquad y=b.$$

We now translate the earlier condition $$\dfrac{2}{a}+\dfrac{3}{b}=1$$ into $$x$$ and $$y$$. Replacing $$a$$ by $$x$$ and $$b$$ by $$y$$, we have

$$\frac{2}{x}+\frac{3}{y}=1.$$

To clear the denominators we multiply every term by $$xy$$, obtaining

$$2y+3x=xy.$$

Re-arranging the terms to match the standard presentation, we write

$$3x+2y=xy.$$

This equation represents the path traced by the point $$R$$ for every admissible position of the line through $$(2,3)$$ that meets both coordinate axes.

Hence, the correct answer is Option D.