1. Interpret the given pair of lines

The homogeneous second-degree equation $a x^2 + 2(a+b)xy + b y^2 = 0$ represents a pair of straight lines through the origin.

Since these lines lie along diameters of a circle, they divide the circle into four sectors. If the angle between the two lines is $\theta$, then the four sectors have angles: $\theta,\; \pi-\theta,\; \theta,\; \pi-\theta$ So their areas are proportional to these central angles.

Given that one sector has area thrice another, we must have $\pi - \theta = 3\theta$ or equivalently $\theta = 3(\pi-\theta).$ Both give the same acute angle relation: $\theta = \frac{\pi}{4} \quad \text{or} \quad \theta = \frac{3\pi}{4}.$ Thus, the angle between the lines must satisfy $\tan \theta = \pm 1.$

2. Angle between the pair of lines

For a pair of lines $Ax^2 + 2Hxy + By^2 = 0,$ the angle $\theta$ between them is given by $\tan \theta = \frac{2\sqrt{H^2-AB}}{A+B}$ (when defined).

Here, $A=a,\quad H=a+b,\quad B=b.$ So $H^2-AB = (a+b)^2-ab = a^2+ab+b^2.$ Hence $\tan \theta = \frac{2\sqrt{a^2+ab+b^2}}{a+b}.$

Since the sector condition gives $\tan\theta = \pm 1$, we impose $\left(\frac{2\sqrt{a^2+ab+b^2}}{a+b}\right)^2 = 1.$ Therefore, $4(a^2+ab+b^2) = (a+b)^2.$ Expanding, $4a^2+4ab+4b^2 = a^2+2ab+b^2.$ So, $3a^2+2ab+3b^2 = 0.$

3. Match with the options

This corresponds to: $\boxed{3a^2 + 2ab + 3b^2 = 0}$ which is Option D.

4. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

So the derived answer agrees with the stored correct answer.