A line passing through the point P(a, $$\theta$$) makes an acute angle $$\alpha$$ with the positive x-axis. Let this line be rotated about the point P through an angle $$\frac{\alpha}{2}$$ in the clock-wise direction. If in the new position, the slope of the line is $$2 - \sqrt{3}$$ and its distance from the origin is $$\frac{1}{\sqrt{2}}$$, then the value of $$3a^2\tan^2\alpha - 2\sqrt{3}$$ is

The first position of the line is through the point $$P(a,0)$$ and makes an acute angle $$\alpha$$ with the positive $$x$$-axis, so its slope is $$m_1=\tan\alpha$$.

The line is then rotated clockwise about $$P$$ through $$\dfrac{\alpha}{2}$$.
Clockwise rotation decreases the inclination, hence the new angle with the $$x$$-axis is $$\alpha-\dfrac{\alpha}{2}= \dfrac{\alpha}{2}$$.
Therefore the slope of the new line is $$m_2=\tan\!\left(\dfrac{\alpha}{2}\right)$$.

Given that in this new position the slope equals $$2-\sqrt{3}$$, we have
$$\tan\!\left(\dfrac{\alpha}{2}\right)=2-\sqrt{3}$$ $$-(1)$$

Formula used: the double-angle identity for tangent,
$$\tan\alpha=\dfrac{2\tan\!\left(\dfrac{\alpha}{2}\right)}{1-\tan^2\!\left(\dfrac{\alpha}{2}\right)}$$ $$-(2)$$

Substituting $$\tan\!\left(\dfrac{\alpha}{2}\right)=2-\sqrt{3}$$ into $$(2)$$:
Numerator $$=2(2-\sqrt{3})=4-2\sqrt{3}$$
Denominator $$=1-(2-\sqrt{3})^{2}=1-\left(7-4\sqrt{3}\right)=4\sqrt{3}-6$$.

Thus
$$\tan\alpha=\dfrac{4-2\sqrt{3}}{4\sqrt{3}-6} =\dfrac{2-\sqrt{3}}{\sqrt{3}(2-\sqrt{3})} =\dfrac{1}{\sqrt{3}}$$.

Since $$\alpha$$ is acute, $$\alpha=30^{\circ}$$ and
$$\tan^{2}\alpha=\dfrac{1}{3}$$ $$-(3)$$

Equation of the new line (slope $$2-\sqrt{3}$$, passing through $$P(a,0)$$):
$$y=(2-\sqrt{3})(x-a)\;\;\Longrightarrow\;\;(2-\sqrt{3})x - y - a(2-\sqrt{3})=0$$.

For a line $$Ax+By+C=0$$, the perpendicular distance from the origin $$(0,0)$$ is
$$d=\dfrac{|C|}{\sqrt{A^{2}+B^{2}}}$$.

Here $$A=2-\sqrt{3},\,B=-1,\,C=-a(2-\sqrt{3})$$, so
$$d=\dfrac{|a|(2-\sqrt{3})}{\sqrt{(2-\sqrt{3})^{2}+1}}$$.

Given distance $$d=\dfrac{1}{\sqrt{2}}$$, we get
$$\dfrac{a^{2}(2-\sqrt{3})^{2}}{(2-\sqrt{3})^{2}+1}=\dfrac{1}{2}$$.

Let $$t=2-\sqrt{3}\;(\,t^{2}=7-4\sqrt{3}\,)$$. Then
$$a^{2}=\dfrac{\dfrac12\bigl(1+t^{2}\bigr)}{t^{2}} =\dfrac12\cdot\dfrac{1+7-4\sqrt{3}}{t^{2}} =\dfrac12\cdot\dfrac{8-4\sqrt{3}}{t^{2}} =\dfrac{2(2-\sqrt{3})}{t^{2}} =\dfrac{2t}{t^{2}} =\dfrac{2}{t}$$.

Since $$t=2-\sqrt{3}$$,
$$a^{2}=\dfrac{2}{2-\sqrt{3}} =2\cdot\dfrac{2+\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})} =2(2+\sqrt{3}) =4+2\sqrt{3}$$ $$-(4)$$

Required expression:
$$3a^{2}\tan^{2}\alpha-2\sqrt{3} =3a^{2}\cdot\dfrac{1}{3}-2\sqrt{3} =a^{2}-2\sqrt{3}$$ (using $$(3)$$).

Using $$(4)$$,
$$a^{2}-2\sqrt{3}=(4+2\sqrt{3})-2\sqrt{3}=4.$$

Therefore, $$3a^{2}\tan^{2}\alpha-2\sqrt{3}=4$$, which corresponds to Option A.