The circumcentre of a triangle lies at the origin and its centroid is the midpoint of the line segment joining the points $$(a^2 + 1, a^2 + 1)$$ and $$(2a, -2a)$$, $$a \neq 0$$. Then for any a, the orthocentre of this triangle lies on the line:

The circumcentre of the triangle is at the origin, denoted as $$O(0, 0)$$. The centroid $$G$$ is given as the midpoint of the line segment joining the points $$(a^2 + 1, a^2 + 1)$$ and $$(2a, -2a)$$, where $$a \neq 0$$.

Using the midpoint formula, the coordinates of the centroid $$G$$ are calculated as follows:

$$$ x_G = \frac{(a^2 + 1) + (2a)}{2} = \frac{a^2 + 2a + 1}{2} $$$

$$$ y_G = \frac{(a^2 + 1) + (-2a)}{2} = \frac{a^2 - 2a + 1}{2} $$$

Simplifying the expressions:

$$$ a^2 + 2a + 1 = (a + 1)^2 $$$

$$$ a^2 - 2a + 1 = (a - 1)^2 $$$

So,

$$$ x_G = \frac{(a + 1)^2}{2} $$$

$$$ y_G = \frac{(a - 1)^2}{2} $$$

In any triangle, the centroid divides the line segment joining the orthocentre $$H$$ and circumcentre $$O$$ in the ratio $$2:1$$, with the centroid closer to the orthocentre. The vector relation is:

$$$ \overrightarrow{G} = \frac{2\overrightarrow{O} + \overrightarrow{H}}{3} $$$

Since the circumcentre is at the origin, $$\overrightarrow{O} = \vec{0}$$, so:

$$$ \overrightarrow{G} = \frac{\overrightarrow{H}}{3} $$$

Thus,

$$$ \overrightarrow{H} = 3\overrightarrow{G} $$$

Therefore, the coordinates of the orthocentre $$H$$ are three times the coordinates of the centroid $$G$$:

$$$ x_H = 3 \times x_G = 3 \times \frac{(a + 1)^2}{2} = \frac{3(a + 1)^2}{2} $$$

$$$ y_H = 3 \times y_G = 3 \times \frac{(a - 1)^2}{2} = \frac{3(a - 1)^2}{2} $$$

So, $$H$$ is at $$\left( \frac{3(a + 1)^2}{2}, \frac{3(a - 1)^2}{2} \right)$$.

The problem requires finding the line on which $$H$$ lies for any $$a \neq 0$$. Substituting the coordinates of $$H$$ into each option will determine which equation is satisfied identically for all $$a$$.

Option A: $$y - (a^2 + 1)x = 0$$

Substitute $$x_H$$ and $$y_H$$:

$$$ y_H - (a^2 + 1)x_H = \frac{3(a - 1)^2}{2} - (a^2 + 1) \cdot \frac{3(a + 1)^2}{2} = \frac{3}{2} \left[ (a - 1)^2 - (a^2 + 1)(a + 1)^2 \right] $$$

Expand:

$$$ (a - 1)^2 = a^2 - 2a + 1 $$$

$$$ (a + 1)^2 = a^2 + 2a + 1 $$$

$$$ (a^2 + 1)(a + 1)^2 = (a^2 + 1)(a^2 + 2a + 1) = a^4 + 2a^3 + a^2 + a^2 + 2a + 1 = a^4 + 2a^3 + 2a^2 + 2a + 1 $$$

So,

$$$ (a - 1)^2 - (a^2 + 1)(a + 1)^2 = (a^2 - 2a + 1) - (a^4 + 2a^3 + 2a^2 + 2a + 1) = -a^4 - 2a^3 - a^2 - 4a $$$

This is not zero for all $$a$$, so Option A is incorrect.

Option B: $$y - 2a x = 0$$

Substitute $$x_H$$ and $$y_H$$:

$$$ y_H - 2a x_H = \frac{3(a - 1)^2}{2} - 2a \cdot \frac{3(a + 1)^2}{2} = \frac{3}{2} \left[ (a - 1)^2 - 2a (a + 1)^2 \right] $$$

Expand:

$$$ (a - 1)^2 = a^2 - 2a + 1 $$$

$$$ 2a (a + 1)^2 = 2a (a^2 + 2a + 1) = 2a^3 + 4a^2 + 2a $$$

So,

$$$ (a - 1)^2 - 2a (a + 1)^2 = (a^2 - 2a + 1) - (2a^3 + 4a^2 + 2a) = -2a^3 - 3a^2 - 4a + 1 $$$

This is not zero for all $$a$$, so Option B is incorrect.

Option C: $$y + x = 0$$

Substitute $$x_H$$ and $$y_H$$:

$$$ y_H + x_H = \frac{3(a - 1)^2}{2} + \frac{3(a + 1)^2}{2} = \frac{3}{2} \left[ (a - 1)^2 + (a + 1)^2 \right] $$$

Expand:

$$$ (a - 1)^2 + (a + 1)^2 = (a^2 - 2a + 1) + (a^2 + 2a + 1) = 2a^2 + 2 $$$

So,

$$$ \frac{3}{2} \times (2a^2 + 2) = 3(a^2 + 1) $$$

This equals zero only if $$a^2 + 1 = 0$$, which is not true for real $$a$$, so Option C is incorrect.

Option D: $$(a - 1)^2 x - (a + 1)^2 y = 0$$

Substitute $$x_H$$ and $$y_H$$:

$$$ (a - 1)^2 x_H - (a + 1)^2 y_H = (a - 1)^2 \cdot \frac{3(a + 1)^2}{2} - (a + 1)^2 \cdot \frac{3(a - 1)^2}{2} $$$

Factor out $$\frac{3}{2}$$:

$$$ = \frac{3}{2} (a - 1)^2 (a + 1)^2 - \frac{3}{2} (a + 1)^2 (a - 1)^2 = 0 $$$

This is identically zero for all $$a \neq 0$$, so the orthocentre $$H$$ lies on this line for any $$a$$.

Hence, the correct answer is Option D.