Let ` sqrt(3) hati + hatj , hati + sqrt(3)hatj` 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 C from the bisector of the acute angle between OA and OB.

### Step-by-Step Solution:

1. **Identify the Position Vectors**:
- The position vector of point A is given as \( \vec{A} = \sqrt{3} \hat{i} + \hat{j} \).
- The position vector of point B is given as \( \vec{B} = \hat{i} + \sqrt{3} \hat{j} \).
- The position vector of point C is given as \( \vec{C} = \beta \hat{i} + (1 - \beta) \hat{j} \).

2. **Determine the Angles**:
- The angle \( \theta_A \) for vector OA is given by:
\[ \tan \theta_A = \frac{1}{\sqrt{3}} \implies \theta_A = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]
- The angle \( \theta_B \) for vector OB is given by:
\[ \tan \theta_B = \sqrt{3} \implies \theta_B = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \]

3. **Find the Angle Bisector**:
- The angle bisector of the acute angle between OA and OB is given by:
\[ \theta_{bisector} = \frac{\theta_A + \theta_B}{2} = \frac{\frac{\pi}{6} + \frac{\pi}{3}}{2} = \frac{\pi}{4} \]
- The slope of the bisector is:
\[ \tan\left(\frac{\pi}{4}\right) = 1 \]
- Therefore, the equation of the bisector (which passes through the origin) is:
\[ y = x \quad \text{or} \quad x - y = 0 \]

4. **Distance from Point C to the Bisector**:
- The distance \( d \) from point C to the line \( x - y = 0 \) is given by the formula:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
- Here, \( A = 1, B = -1, C = 0 \), and \( (x_1, y_1) = (\beta, 1 - \beta) \):
\[ d = \frac{|1 \cdot \beta - 1 \cdot (1 - \beta) + 0|}{\sqrt{1^2 + (-1)^2}} = \frac{|\beta - (1 - \beta)|}{\sqrt{2}} = \frac{|2\beta - 1|}{\sqrt{2}} \]

5. **Set the Distance Equal to Given Value**:
- We know from the problem that this distance is \( \frac{3}{\sqrt{2}} \):
\[ \frac{|2\beta - 1|}{\sqrt{2}} = \frac{3}{\sqrt{2}} \]
- Cross-multiplying gives:
\[ |2\beta - 1| = 3 \]

6. **Solve the Absolute Value Equation**:
- This leads to 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**:
\[ 2\beta - 1 = -3 \implies 2\beta = -2 \implies \beta = -1 \]

7. **Sum of All Possible Values of \( \beta \)**:
- The possible values of \( \beta \) are \( 2 \) and \( -1 \).
- Therefore, the sum is:
\[ 2 + (-1) = 1 \]

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