Let ` sqrt(3i) + hatj , hati + sqrt(3j) and beta hati + (1- beta)hatj` respectively be the position vedors of the points A, B and C with respect the origin O. If the distance of C from the bisector of the acute angle between OA and OB is `(3)/(sqrt2),` then the sum all possible values of `beta` is `"_______."`

Explanation

To solve the problem, we need to find the sum of all possible values of \( \beta \) given the position vectors of points A, B, and C, and the distance of point C from the bisector of the acute angle between OA and OB.

### Step 1: Identify the position vectors
The position vectors of points A, B, and C are given as:
- \( \vec{OA} = \sqrt{3} \hat{i} + \hat{j} \)
- \( \vec{OB} = \hat{i} + \sqrt{3} \hat{j} \)
- \( \vec{OC} = \beta \hat{i} + (1 - \beta) \hat{j} \)

### Step 2: Convert position vectors to coordinates
The coordinates of points A, B, and C can be expressed as:
- Point A: \( ( \sqrt{3}, 1 ) \)
- Point B: \( ( 1, \sqrt{3} ) \)
- Point C: \( ( \beta, 1 - \beta ) \)

### Step 3: Find the slopes of OA and OB
The slope of line OA (from origin O to point A) is:
\[ \text{slope of OA} = \frac{1 - 0}{\sqrt{3} - 0} = \frac{1}{\sqrt{3}} \quad \text{(30 degrees)} \]

The slope of line OB (from origin O to point B) is:
\[ \text{slope of OB} = \frac{\sqrt{3} - 0}{1 - 0} = \sqrt{3} \quad \text{(60 degrees)} \]

### Step 4: Find the angle bisector
The angle bisector of the acute angle between OA and OB will have a slope that is the average of the slopes of OA and OB. The angle bisector will make an angle of \( 45^\circ \) with the x-axis, so its equation is:
\[ y = x \]
or equivalently,
\[ y - x = 0 \]

### Step 5: Calculate the distance from point C to the bisector
The distance \( D \) from a point \( (x_1, y_1) \) to the line \( Ax + By + C = 0 \) is given by:
\[ D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For the line \( y - x = 0 \) (which can be rewritten as \( 1 \cdot x - 1 \cdot y + 0 = 0 \)), we have:
- \( A = 1 \)
- \( B = -1 \)
- \( C = 0 \)

Substituting \( (x_1, y_1) = (\beta, 1 - \beta) \):
\[ D = \frac{|1 \cdot \beta - 1 \cdot (1 - \beta) + 0|}{\sqrt{1^2 + (-1)^2}} = \frac{|\beta - (1 - \beta)|}{\sqrt{2}} = \frac{|2\beta - 1|}{\sqrt{2}} \]

### Step 6: Set the distance equal to the given distance
We are given that the distance \( D \) is \( \frac{3}{\sqrt{2}} \):
\[ \frac{|2\beta - 1|}{\sqrt{2}} = \frac{3}{\sqrt{2}} \]
Multiplying both sides by \( \sqrt{2} \):
\[ |2\beta - 1| = 3 \]

### Step 7: Solve the absolute value equation
This gives us two cases:
1. \( 2\beta - 1 = 3 \)
2. \( 2\beta - 1 = -3 \)

**Case 1:**
\[ 2\beta - 1 = 3 \implies 2\beta = 4 \implies \beta = 2 \]

**Case 2:**
\[ 2\beta - 1 = -3 \implies 2\beta = -2 \implies \beta = -1 \]

### Step 8: Sum all possible values of \( \beta \)
The possible values of \( \beta \) are \( 2 \) and \( -1 \). Therefore, the sum of all possible values of \( \beta \) is:
\[ 2 + (-1) = 1 \]

### Final Answer
The sum of all possible values of \( \beta \) is:
\[ \boxed{1} \]