Let A, B, C be three points in $$xy-plane$$, whose position vector are given by $$\sqrt{3}\hat{i}+\hat{j}, \hat{i}+\sqrt{3}\hat{j}$$ and $$a\hat{i}+ (1-a)\hat{j}$$ respectively with respect to the origin O . If the distance of the point C from the line bisecting the angle between the vectors $$\overrightarrow{OA}$$ and $$\overrightarrow{OB}$$ is $$\frac{9}{\sqrt{2}}$$, then the sum of all the possible values of $$a$$ is :

Let the position vectors of points A, B and C be given by $$\overrightarrow{OA}=\sqrt{3}\,\hat{i}+\hat{j}$$, $$\overrightarrow{OB}=\hat{i}+\sqrt{3}\,\hat{j}$$ and $$\overrightarrow{OC}=a\,\hat{i}+(1-a)\,\hat{j}$$ respectively.

We first find the equation of the internal angle bisector of the angle between $$\overrightarrow{OA}$$ and $$\overrightarrow{OB}$$.

Formula for internal bisector direction: if $$\mathbf{v}_1$$ and $$\mathbf{v}_2$$ are two vectors, then a direction vector along the internal bisector is $$\mathbf{u}=\frac{\mathbf{v}_1}{\|\mathbf{v}_1\|}+\frac{\mathbf{v}_2}{\|\mathbf{v}_2\|}$$.

Here, $$\|\overrightarrow{OA}\|=\sqrt{(\sqrt{3})^2+1^2}=\sqrt{3+1}=2$$ and $$\|\overrightarrow{OB}\|=\sqrt{1^2+(\sqrt{3})^2}=2$$.

Thus $$\mathbf{u} =\frac{1}{2}(\sqrt{3},1)+\frac{1}{2}(1,\sqrt{3}) =\Bigl(\frac{\sqrt{3}+1}{2},\frac{1+\sqrt{3}}{2}\Bigr) \;\Longrightarrow\; \text{direction }(d_x,d_y)=(\sqrt{3}+1,\;1+\sqrt{3}).$$

Since $$\sqrt{3}+1=1+\sqrt{3}$$, the slope of this line is $$m=\frac{1+\sqrt{3}}{\sqrt{3}+1}=1.$$ Hence the internal bisector is the line $$y-x=0\quad\text{or}\quad x-y=0.$$

Point C has coordinates $$(a,\;1-a)$$. The distance of $$(a,1-a)$$ from the line $$x-y=0$$ is given by the formula $$\text{Distance}=\frac{\bigl|a-(1-a)\bigr|}{\sqrt{1^2+(-1)^2}} =\frac{|2a-1|}{\sqrt{2}}\quad-(1)$$.

We are given that this distance equals $$\dfrac{9}{\sqrt{2}}$$. Equating with $$(1)$$ we get: $$\frac{|2a-1|}{\sqrt{2}}=\frac{9}{\sqrt{2}} \;\Longrightarrow\;|2a-1|=9.$$ This gives two cases:

Case 1: $$2a-1=9\quad\Longrightarrow\quad2a=10\quad\Longrightarrow\quad a=5.$$

Case 2: $$2a-1=-9\quad\Longrightarrow\quad2a=-8\quad\Longrightarrow\quad a=-4.$$

Thus the possible values of $$a$$ are $$5$$ and $$-4$$. Their sum is $$5+(-4)=1.$$

Final Answer: The sum of all possible values of $$a$$ is $$1$$ (Option C).