If the pair of lines $$ax^2 + 2(a + b)xy + by^2 = 0$$ lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then

The equation $$ax^2 + 2(a+b)xy + by^2 = 0$$ represents two straight lines through the origin (centre of the circle).
For a homogeneous second-degree expression $$ax^2 + 2hxy + by^2 = 0$$ the two lines make an angle $$\theta$$ given by

$$\tan\theta = \left|\dfrac{2\sqrt{\,h^2-ab\,}}{a+b}\right| \; -(1)$$

Here $$h = a + b$$, so

$$\tan\theta = \left|\dfrac{2\sqrt{(a+b)^2 - ab}}{a + b}\right| = \left|\dfrac{2\sqrt{a^2 + ab + b^2}}{a + b}\right| \; -(2)$$

Because the two lines are diameters, they divide the circle into four sectors whose central angles are the angles between the lines and their supplements. Let the smaller angle between the lines be $$\theta$$; the opposite angle is also $$\theta$$, while the other two are $$\pi - \theta$$ each.

One sector has area thrice that of another, so the corresponding central angles are in the ratio $$3:1$$.
Let the smaller sector have angle $$\theta$$ and the larger sector have angle $$\pi - \theta$$. Then

$$(\pi - \theta) = 3\theta \;\;\Longrightarrow\;\; \pi = 4\theta \;\;\Longrightarrow\;\; \theta = \dfrac{\pi}{4}$$

Hence $$\theta = 45^\circ$$ and therefore $$\tan\theta = 1$$.

Substituting $$\tan\theta = 1$$ in equation $$-(2)$$:

$$\dfrac{2\sqrt{a^2 + ab + b^2}}{|a + b|} = 1$$

Square both sides:

$$4\left(a^2 + ab + b^2\right) = (a + b)^2$$

Expand and collect terms:

$$4a^2 + 4ab + 4b^2 - (a^2 + 2ab + b^2) = 0$$
$$3a^2 + 2ab + 3b^2 = 0$$

This is exactly the condition given in Option D.

Final Answer: Option D which is: $$3a^2 + 2ab + 3b^2 = 0$$