Let the point A divide the line segment joining the points P(−1,−1, 2) and Q(5, 5, 10) internally in the ratio $$r : 1 (r > 0)$$. If O is the origin and $$(\overrightarrow{OQ}.\overrightarrow{OA})-\frac{1}{5}|\overrightarrow{OP}.\overrightarrow{OA}|^{2}=10$$. then the value of r is :

We are given $$P(-1, -1, 2)$$, $$Q(5, 5, 10)$$, and $$O$$ is the origin.

Point $$A$$ divides $$PQ$$ internally in the ratio $$r : 1$$. By the section formula:

$$\overrightarrow{OA} = \frac{r \cdot \overrightarrow{OQ} + 1 \cdot \overrightarrow{OP}}{r + 1} = \frac{r(5, 5, 10) + (-1, -1, 2)}{r + 1} = \left(\frac{5r - 1}{r + 1}, \frac{5r - 1}{r + 1}, \frac{10r + 2}{r + 1}\right)$$

We have $$\overrightarrow{OQ} = (5, 5, 10)$$ and $$\overrightarrow{OP} = (-1, -1, 2)$$.

Computing $$\overrightarrow{OQ} \cdot \overrightarrow{OA}$$:

$$\overrightarrow{OQ} \cdot \overrightarrow{OA} = \frac{5(5r-1) + 5(5r-1) + 10(10r+2)}{r+1} = \frac{25r - 5 + 25r - 5 + 100r + 20}{r+1} = \frac{150r + 10}{r+1}$$

The problem involves $$|\overrightarrow{OP} \times \overrightarrow{OA}|^2$$ (cross product). Computing $$\overrightarrow{OP} \times \overrightarrow{OA}$$:

$$\overrightarrow{OP} \times \overrightarrow{OA} = \frac{1}{r+1} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & 2 \\ 5r-1 & 5r-1 & 10r+2 \end{vmatrix}$$

$$\hat{i}$$-component: $$(-1)(10r+2) - 2(5r-1) = -10r - 2 - 10r + 2 = -20r$$

$$\hat{j}$$-component: $$-((-1)(10r+2) - 2(5r-1)) = -(-10r - 2 - 10r + 2) = 20r$$

$$\hat{k}$$-component: $$(-1)(5r-1) - (-1)(5r-1) = 0$$

$$\overrightarrow{OP} \times \overrightarrow{OA} = \frac{1}{r+1}(-20r, 20r, 0) = \frac{20r}{r+1}(-1, 1, 0)$$

$$|\overrightarrow{OP} \times \overrightarrow{OA}|^2 = \frac{400r^2}{(r+1)^2} \times (1 + 1 + 0) = \frac{800r^2}{(r+1)^2}$$

Substituting into the given equation $$(\overrightarrow{OQ} \cdot \overrightarrow{OA}) - \frac{1}{5}|\overrightarrow{OP} \times \overrightarrow{OA}|^2 = 10$$:

$$\frac{150r + 10}{r+1} - \frac{1}{5} \cdot \frac{800r^2}{(r+1)^2} = 10$$

$$\frac{150r + 10}{r+1} - \frac{160r^2}{(r+1)^2} = 10$$

Multiplying through by $$(r+1)^2$$:

$$(150r + 10)(r+1) - 160r^2 = 10(r+1)^2$$

$$150r^2 + 150r + 10r + 10 - 160r^2 = 10r^2 + 20r + 10$$

$$-10r^2 + 160r + 10 = 10r^2 + 20r + 10$$

$$0 = 20r^2 - 140r$$

$$0 = 20r(r - 7)$$

Since $$r > 0$$, we get $$r = 7$$.

The answer is Option D: $$7$$.