1. Write the coordinates of the points

The given position vectors are:

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

So,

$A(\sqrt{3},1),\qquad B(1,\sqrt{3}),\qquad C(\beta,1-\beta).$

2. Find the bisector of the acute angle between $OA$ and $OB$

The vectors $\vec{OA}$ and $\vec{OB}$ have equal magnitudes:

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

Hence the internal angle bisector is along

$\frac{\vec{OA}}{|\vec{OA}|}+\frac{\vec{OB}}{|\vec{OB}|}$

Since both magnitudes are equal, this is just along

$\vec{OA}+\vec{OB}=(\sqrt{3}+1)\hat i+(\sqrt{3}+1)\hat j.$

So the bisector has direction $(1,1)$, hence its equation is

$y=x.$

3. Use the distance of point $C$ from the line $y=x$

Line $y=x$ can be written as

$x-y=0.$

Distance of point $(x_1,y_1)$ from line $ax+by+c=0$ is

$d=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}.$

Here, for $C(\beta,1-\beta)$,

$d=\frac{|\beta-(1-\beta)|}{\sqrt{1^2+(-1)^2}}=\frac{|2\beta-1|}{\sqrt{2}}.$

Given that this distance is

$\frac{3}{\sqrt{2}},$

so

$\frac{|2\beta-1|}{\sqrt{2}}=\frac{3}{\sqrt{2}}.$

Therefore,

$|2\beta-1|=3.$

So,

$2\beta-1=3 \quad \text{or} \quad 2\beta-1=-3.$

Thus,

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

4. Find the sum of all possible values of $\beta$

$2+(-1)=1.$

5. Check with options

The correct option is:

$\boxed{\text{B: }1}$

6. Comparison with stored answer

Stored correct answer = B

Our derived answer = B

So they agree.