Let $$PS$$ be the median of the triangle with vertices $$P(2, 2)$$, $$Q(6, -1)$$ and $$R(7, 3)$$. The equation of the line passing through $$(1, -1)$$ and parallel to $$PS$$ is:

We have the three vertices of the triangle as $$P(2,\,2),\;Q(6,\,-1)\; \text{and}\; R(7,\,3)$$. The median $$PS$$ is drawn from the vertex $$P$$ to the midpoint $$S$$ of the side $$QR$$, so we first find the coordinates of this midpoint.

The midpoint formula says that the midpoint of $$\bigl(x_1,\,y_1\bigr)$$ and $$\bigl(x_2,\,y_2\bigr)$$ is

$$\left(\dfrac{x_1+x_2}{2},\;\dfrac{y_1+y_2}{2}\right).$$

Applying the formula to $$Q(6,\,-1)$$ and $$R(7,\,3)$$, we obtain

$$S\;=\;\left(\dfrac{6+7}{2},\;\dfrac{-1+3}{2}\right) \;=\;\left(\dfrac{13}{2},\;\dfrac{2}{2}\right) \;=\;\left(\dfrac{13}{2},\;1\right).$$

Now we determine the slope of the median $$PS$$. The slope formula for the line through points $$\bigl(x_1,\,y_1\bigr)$$ and $$\bigl(x_2,\,y_2\bigr)$$ is

$$m=\dfrac{y_2-y_1}{x_2-x_1}.$$

For the points $$P(2,\,2)$$ and $$S\!\left(\dfrac{13}{2},\,1\right)$$, we have

$$m_{PS} =\dfrac{1-2}{\dfrac{13}{2}-2} =\dfrac{-1}{\dfrac{13}{2}-\dfrac{4}{2}} =\dfrac{-1}{\dfrac{9}{2}} =-\dfrac{1}{9/2} =-\dfrac{2}{9}.$$

Hence the slope of any line parallel to $$PS$$ is also $$-\dfrac{2}{9}$$. We now need the equation of the line that passes through the point $$(1,\,-1)$$ and has this slope.

We use the point-slope form of a straight line, which is

$$y-y_1=m\bigl(x-x_1\bigr),$$

where $$\bigl(x_1,\,y_1\bigr)=(1,\,-1)$$ and $$m=-\dfrac{2}{9}$$. Substituting,

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

Simplifying step by step:

$$y+1=-\dfrac{2}{9}x+\dfrac{2}{9}.$$

To clear the denominator, multiply every term by $$9$$:

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

Expanding the left side gives

$$9y+9=-2x+2.$$

Now bring every term to the same side to get the equation in standard form:

$$2x+9y+9-2=0 \;\Longrightarrow\; 2x+9y+7=0.$$

Thus the required equation is

$$2x+9y+7=0.$$

This matches Option D in the given list. Hence, the correct answer is Option D.