A line passing through the point $$A(9, 0)$$ makes an angle of $$30°$$ with the positive direction of $$x$$-axis. If this line is rotated about $$A$$ through an angle of $$15°$$ in the clockwise direction, then its equation in the new position is

The original line passes through $$A(9, 0)$$ and makes an angle of $$30°$$ with the positive x-axis. After a clockwise rotation of $$15°$$, the new angle is $$30° - 15° = 15°$$.

The slope of the rotated line is $$m = \tan 15°$$, and using the identity $$\tan 15° = \tan(45° - 30°) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3}-1)^2}{2} = 2 - \sqrt{3}$$.

Through $$(9, 0)$$ with slope $$2 - \sqrt{3}$$, the line equation is $$ y - 0 = (2 - \sqrt{3})(x - 9), $$ so $$ y = (2 - \sqrt{3})x - 9(2 - \sqrt{3}). $$

Rearranging gives $$y - (2 - \sqrt{3})x = -9(2 - \sqrt{3}),$$ or $$(2 - \sqrt{3})x - y = 9(2 - \sqrt{3}).$$ Dividing by $$(2 - \sqrt{3})$$ yields $$x - \frac{y}{2 - \sqrt{3}} = 9.$$

Note that $$\frac{1}{2 - \sqrt{3}} = \frac{2 + \sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})} = \frac{2+\sqrt{3}}{1} = 2 + \sqrt{3}$$.

Rewriting in terms of $$(\sqrt{3}-2)$$ gives $$\frac{y}{\sqrt{3} - 2} + x = 9,$$ which is equivalently $$x + \frac{y}{\sqrt{3}-2} = 9.$$

The answer is Option (1): $$\boxed{\frac{y}{\sqrt{3}-2} + x = 9}$$.