Let $$D$$ be the centroid of the triangle with vertices $$(3, -1)$$, $$(1, 3)$$ and $$(2, 4)$$. Let P be the point of intersection of the lines $$x + 3y - 1 = 0$$ and $$3x - y + 1 = 0$$. Then, the line passing through the points $$D$$ and $$P$$ also passes through the point:

We have the three vertices of the triangle as $$A(3,-1),\;B(1,3),\;C(2,4)$$. The centroid of a triangle with vertices $$\bigl(x_1,y_1\bigr),\;\bigl(x_2,y_2\bigr),\;\bigl(x_3,y_3\bigr)$$ is given by the well-known formula

$$\Bigl(\dfrac{x_1+x_2+x_3}{3},\;\dfrac{y_1+y_2+y_3}{3}\Bigr).$$

Substituting the coordinates of $$A,B,C$$ we obtain

$$x\text{-coordinate of }D=\dfrac{3+1+2}{3}=\dfrac{6}{3}=2,$$ $$y\text{-coordinate of }D=\dfrac{-1+3+4}{3}=\dfrac{6}{3}=2.$$

So the centroid is $$D(2,2).$$

Next we determine the point $$P$$ where the two straight lines intersect. The equations of the lines are

$$x+3y-1=0\quad\text{and}\quad3x-y+1=0.$$

From the first equation we isolate $$x$$:

$$x = 1 - 3y.$$

Now we substitute this expression for $$x$$ in the second equation:

$$3(1-3y) - y + 1 = 0.$$

Simplifying step by step,

$$3 - 9y - y + 1 = 0,$$ $$4 - 10y = 0,$$ $$10y = 4,$$ $$y = \dfrac{4}{10} = \dfrac{2}{5}.$$

Substituting $$y=\dfrac{2}{5}$$ back into $$x = 1 - 3y$$ gives

$$x = 1 - 3\left(\dfrac{2}{5}\right) = 1 - \dfrac{6}{5} = -\,\dfrac{1}{5}.$$

Thus the intersection point is $$P\!\left(-\dfrac{1}{5},\,\dfrac{2}{5}\right).$$

We now require the equation of the straight line passing through $$D(2,2)$$ and $$P\!\left(-\dfrac{1}{5},\,\dfrac{2}{5}\right).$$ The slope (gradient) of a line through two points $$\bigl(x_1,y_1\bigr)$$ and $$\bigl(x_2,y_2\bigr)$$ is given by

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

So,

$$m=\dfrac{2-\dfrac{2}{5}}{2-\left(-\dfrac{1}{5}\right)}=\dfrac{\dfrac{10}{5}-\dfrac{2}{5}}{\dfrac{10}{5}+\dfrac{1}{5}}=\dfrac{\dfrac{8}{5}}{\dfrac{11}{5}}=\dfrac{8}{11}.$$

Using the point-slope form $$y-y_1=m(x-x_1)$$ with the point $$D(2,2)$$, we write

$$y-2=\dfrac{8}{11}\,\bigl(x-2\bigr).$$

To clear the denominator, multiply every term by $$11$$:

$$11(y-2)=8(x-2).$$

Expanding the brackets,

$$11y-22=8x-16.$$

Bringing all terms to the left side,

$$8x-11y+6=0.$$

This is the equation of the line through $$D$$ and $$P$$. Now we test which of the given option points satisfies $$8x-11y+6=0$$.

Option A: $$(-9,-6)$$ Substituting, $$8(-9)-11(-6)+6=-72+66+6=0.$$ So $$(-9,-6)$$ lies on the line.

Option B: $$(9,7)$$ $$8(9)-11(7)+6=72-77+6=1\neq0.$$

Option C: $$(7,6)$$ $$8(7)-11(6)+6=56-66+6=-4\neq0.$$

Option D: $$(-9,-7)$$ $$8(-9)-11(-7)+6=-72+77+6=11\neq0.$$

Only Option A satisfies the line equation. Hence, the correct answer is Option A.