1. Write the coordinates of the points

The given position vectors are:

$\vec{OA}=\sqrt{3}\hat i+\hat j \implies A(\sqrt{3},1)$ $\vec{OB}=\hat i+\sqrt{3}\hat j \implies B(1,\sqrt{3})$ $\vec{OC}=a\hat i+(1-a)\hat j \implies C(a,1-a)$

2. Find the line bisecting the angle between $\vec{OA}$ and $\vec{OB}$

First note that

$|\vec{OA}|=\sqrt{(\sqrt{3})^2+1^2}=2, \qquad |\vec{OB}|=\sqrt{1^2+(\sqrt{3})^2}=2$

So the unit vectors along them are

$\hat u_A=\frac{1}{2}(\sqrt{3},1), \qquad \hat u_B=\frac{1}{2}(1,\sqrt{3})$

The internal angle bisector has direction

$\hat u_A+\hat u_B=\left(\frac{\sqrt{3}+1}{2},\frac{\sqrt{3}+1}{2}\right)$

Hence its direction is proportional to $(1,1)$, so the bisector line is

$y=x$

3. Use the distance of point $C(a,1-a)$ from the line $y=x$

The line $y=x$ can be written as

$x-y=0$

Distance of $(x_1,y_1)$ from $Ax+By+C=0$ is

$d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$

So distance of $C(a,1-a)$ from $x-y=0$ is

$d=\frac{|a-(1-a)|}{\sqrt{2}}=\frac{|2a-1|}{\sqrt{2}}$

Given that this distance is

$\frac{9}{\sqrt{2}}$

Therefore,

$\frac{|2a-1|}{\sqrt{2}}=\frac{9}{\sqrt{2}}$

$|2a-1|=9$

So,

$2a-1=9 \quad \text{or} \quad 2a-1=-9$

This gives

$a=5 \quad \text{or} \quad a=-4$

4. Find the sum of all possible values of $a$

$5+(-4)=1$

5. Check with options

The sum is

$1$

So the correct option is D.

6. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.