If OA is equally inclined to OX,OY ,OZ and if A is `sqrt(3)` units from the origin then the cordinates of A are

Options

A. `(3,3,3)`
B. `(-1,1,-1)`
C. `(-1,1,1)`
D. `(1,1,1)`

Explanation

To solve the problem, we need to find the coordinates of point A given that OA is equally inclined to the OX, OY, and OZ axes, and that the distance OA is \(\sqrt{3}\) units from the origin.

### Step-by-Step Solution:

1. **Understanding the Angles**:
- Since OA is equally inclined to the OX, OY, and OZ axes, let the angle made with each axis be \(\alpha\). Therefore, we have:
\[ \alpha = \beta = \gamma \]

2. **Using the Cosine Rule**:
- According to the cosine rule in three dimensions, we have:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]
- Substituting \(\alpha = \beta = \gamma\), we get:
\[ \cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1 \]
\[ 3 \cos^2 \alpha = 1 \]

3. **Solving for \(\cos \alpha\)**:
- Dividing both sides by 3:
\[ \cos^2 \alpha = \frac{1}{3} \]
- Taking the square root:
\[ \cos \alpha = \frac{1}{\sqrt{3}} \]

4. **Finding the Coordinates**:
- The coordinates of point A can be expressed in terms of the distance OA and the cosine of the angles:
\[ A = (OA \cos \alpha, OA \cos \beta, OA \cos \gamma) \]
- Since \(OA = \sqrt{3}\) and \(\cos \alpha = \cos \beta = \cos \gamma = \frac{1}{\sqrt{3}}\):
\[ A = \left(\sqrt{3} \cdot \frac{1}{\sqrt{3}}, \sqrt{3} \cdot \frac{1}{\sqrt{3}}, \sqrt{3} \cdot \frac{1}{\sqrt{3}}\right) \]
- Simplifying this gives:
\[ A = (1, 1, 1) \]

5. **Conclusion**:
- Therefore, the coordinates of point A are:
\[ A = (1, 1, 1) \]

### Final Answer:
The coordinates of point A are \((1, 1, 1)\).