Let $$P(3\cos\alpha, 2\sin\alpha)$$, $$\alpha \neq 0$$, be a point on the ellipse $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$. Let $$Q$$ be a point on the circle $$x^2 + y^2 - 14x - 14y + 82 = 0$$, and $$R$$ be a point on the line $$x + y = 5$$. such that  the centroid of the $$\triangle PQR$$ is $$\left(2 + \cos\alpha,\; 3 + \frac{2}{3}\sin\alpha\right)$$, then the sum of the ordinates of all possible points $$R$$ is :

The ellipse is $$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$$, so a parametric point on it is given as $$P(3\cos\alpha,\;2\sin\alpha)$$, with $$\alpha\neq0$$.

Let $$Q(x_{2},y_{2})$$ lie on the circle
$$(x-7)^{2}+(y-7)^{2}=16$$  $$-(1)$$
and let $$R(x_{3},y_{3})$$ lie on the straight line
$$x+y=5$$  $$-(2)$$.

The centroid $$G$$ of $$\triangle PQR$$ is given to be
$$G\left(2+\cos\alpha,\;3+\frac23\sin\alpha\right).$$
Using the centroid formula $$G\Bigl(\dfrac{x_{1}+x_{2}+x_{3}}{3},\dfrac{y_{1}+y_{2}+y_{3}}{3}\Bigr)$$, equate the coordinates:

$$\frac{3\cos\alpha+x_{2}+x_{3}}{3}=2+\cos\alpha,$$
$$\frac{2\sin\alpha+y_{2}+y_{3}}{3}=3+\frac23\sin\alpha.$$

Solving each for $$x_{3}$$ and $$y_{3}$$:

$$x_{3}=3(2+\cos\alpha)-(3\cos\alpha+x_{2})=6-x_{2},$$
$$y_{3}=3\!\left(3+\frac23\sin\alpha\right)-(2\sin\alpha+y_{2})=9-y_{2}.$$

Notice that both $$x_{3}$$ and $$y_{3}$$ are independent of $$\alpha$$. Hence

$$R(6-x_{2},\;9-y_{2}).$$

Point $$R$$ must also satisfy the line equation $$(2)$$:

$$(6-x_{2})+(9-y_{2})=5\quad\Longrightarrow\quad x_{2}+y_{2}=10.$$

Therefore, point $$Q(x_{2},y_{2})$$ must satisfy the simultaneous system

$$\begin{cases}(x-7)^{2}+(y-7)^{2}=16,\\ x+y=10.\end{cases}$$

Put $$y=10-x$$ into the circle $$-(1)$$:

$$(x-7)^{2}+\bigl[(10-x)-7\bigr]^{2}=16,$$
$$(x-7)^{2}+(3-x)^{2}=16,$$
$$x^{2}-14x+49+x^{2}-6x+9=16,$$
$$2x^{2}-20x+58=16,$$
$$2x^{2}-20x+42=0,$$
$$x^{2}-10x+21=0,$$
$$(x-3)(x-7)=0.$$

Thus
$$x_{2}=3\;\; \text{or}\;\; x_{2}=7.$$
Correspondingly $$y_{2}=10-x_{2}$$ gives

$$Q_{1}(3,7),\qquad Q_{2}(7,3).$$

Substituting each into $$R(6-x_{2},9-y_{2})$$:

For $$Q_{1}(3,7):\; R_{1}(6-3,\;9-7)=(3,2).$$
For $$Q_{2}(7,3):\; R_{2}(6-7,\;9-3)=(-1,6).$$

Both $$R_{1}$$ and $$R_{2}$$ satisfy the line $$x+y=5$$ as required.

Hence the ordinates (y-coordinates) of the possible points $$R$$ are $$2$$ and $$6$$.
Sum of ordinates = $$2+6=8.$$

Option D which is: $$8$$