The combined equation of the two lines $$ax + by + c = 0$$ and $$a'x + b'y + c' = 0$$ can be written as $$ax + by + ca'x + b'y + c' = 0$$. The equation of the angle bisectors of the lines represented by the equation $$2x^2 + xy - 3y^2 = 0$$ is

We are given the pair of straight lines $$2x^2 + xy - 3y^2 = 0$$.

This is of the form $$ax^2 + 2hxy + by^2 = 0$$, where $$a = 2$$, $$2h = 1$$ (so $$h = \frac{1}{2}$$), and $$b = -3$$.

The equation of the angle bisectors of the pair of lines $$ax^2 + 2hxy + by^2 = 0$$ is given by the standard formula:

$$\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$$

Substituting the values $$a = 2$$, $$b = -3$$, and $$h = \frac{1}{2}$$:

$$\frac{x^2 - y^2}{2 - (-3)} = \frac{xy}{\frac{1}{2}}$$

$$\frac{x^2 - y^2}{5} = 2xy$$

Cross-multiplying:

$$x^2 - y^2 = 10xy$$

Rearranging:

$$x^2 - y^2 - 10xy = 0$$

The correct answer is Option D: $$x^2 - y^2 - 10xy = 0$$.