Let $$\sqrt{3}\hat{i} + \hat{j}$$, $$\hat{i} + \sqrt{3}\hat{j}$$ and $$\beta\hat{i} + (1 - \beta)\hat{j}$$ respectively be the position vectors of the points A, B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is $$\frac{\sqrt{3}}{\sqrt{2}}$$, then the sum of all possible values of $$\beta$$ is:

We have the position vectors of the three points with respect to the origin O

$$\overrightarrow{OA}= \sqrt{3}\,\hat i+\hat j,$$ $$\overrightarrow{OB}= \hat i+\sqrt{3}\,\hat j,$$ $$\overrightarrow{OC}= \beta\,\hat i+(1-\beta)\,\hat j.$$

Writing these vectors in coordinate form, the points are

$$A\equiv(\sqrt3,\,1),$$ $$B\equiv(1,\,\sqrt3),$$ $$C\equiv(\beta,\,1-\beta).$$

First we find the bisector of the acute angle between the rays OA and OB. For two rays whose direction vectors are $$\vec a$$ and $$\vec b,$$ the internal bisector has direction $$\dfrac{\vec a}{|\vec a|}+\dfrac{\vec b}{|\vec b|}\;.$$ Therefore we begin by normalising the two direction vectors.

The magnitudes are

$$|\vec a|=\sqrt{(\sqrt3)^2+1^2}=\sqrt{3+1}=2,$$ $$|\vec b|=\sqrt{1^2+(\sqrt3)^2}=\sqrt{1+3}=2.$$

Hence the unit vectors are

$$\frac{\vec a}{|\vec a|}=\left(\frac{\sqrt3}{2},\;\frac12\right),$$ $$\frac{\vec b}{|\vec b|}=\left(\frac12,\;\frac{\sqrt3}{2}\right).$$

Adding them gives the direction of the acute-angle bisector:

$$\left(\frac{\sqrt3}{2}+\frac12,\;\frac12+\frac{\sqrt3}{2}\right) =\left(\frac{\sqrt3+1}{2},\;\frac{\sqrt3+1}{2}\right).$$

Both coordinates are equal, so the bisector is the straight line through the origin having equation

$$y=x.$$

Now we use the perpendicular-distance formula for a point $$(x_0,y_0)$$ from a straight line $$ax+by+c=0:$$

$$\text{Distance}=\frac{|ax_0+by_0+c|}{\sqrt{a^{2}+b^{2}}}\;.$$

The bisector line $$y=x$$ can be rewritten as $$x-y=0,$$ so $$a=1,\;b=-1,\;c=0.$$ For the point $$C(\beta,\,1-\beta)$$ the perpendicular distance is therefore

$$d=\frac{|1\cdot\beta+(-1)(1-\beta)+0|}{\sqrt{1^{2}+(-1)^{2}}} =\frac{|\,\beta-(1-\beta)\,|}{\sqrt2} =\frac{|\,2\beta-1\,|}{\sqrt2}\;.$$

According to the question this distance equals $$\dfrac{\sqrt3}{\sqrt2},$$ so we equate:

$$\frac{|\,2\beta-1\,|}{\sqrt2}=\frac{\sqrt3}{\sqrt2}.$$

Multiplying by $$\sqrt2$$ on both sides gives

$$|\,2\beta-1\,|=\sqrt3.$$

Removing the absolute value produces two linear equations:

$$2\beta-1=\sqrt3\quad\text{or}\quad 2\beta-1=-\sqrt3.$$

Solving each in turn,

$$\begin{aligned} 2\beta-1&=\sqrt3 &\Longrightarrow&\; 2\beta=1+\sqrt3 &\Longrightarrow&\; \beta=\dfrac{1+\sqrt3}{2},\\[4pt] 2\beta-1&=-\sqrt3 &\Longrightarrow&\; 2\beta=1-\sqrt3 &\Longrightarrow&\; \beta=\dfrac{1-\sqrt3}{2}. \end{aligned}$$

Thus the possible values of $$\beta$$ are

$$\beta_1=\frac{1+\sqrt3}{2},\qquad \beta_2=\frac{1-\sqrt3}{2}.$$

The required sum is therefore

$$\beta_1+\beta_2 =\frac{1+\sqrt3}{2}+\frac{1-\sqrt3}{2} =\frac{1+\sqrt3+1-\sqrt3}{2} =\frac{2}{2}=1.$$

Hence, the correct answer is Option D.