If A, B, C and D are four points with position vectors `3hati, 3hatj, 3hatk` and `hati+hatj+hatk` respectively, then D is the

Options

A. orthocentre of `DeltaABC`
B. circumcentre of `DeltaABC`
C. centroid of `DeltaABC`
D. incentre of `DeltaABC`

Answer

To solve the problem, we need to analyze the position vectors of points A, B, C, and D, and determine the properties of point D based on the given vectors.

### Step-by-Step Solution:

1. **Identify the Position Vectors:**
- Let the position vector of point A be \( \vec{A} = 3\hat{i} + 3\hat{j} + 3\hat{k} \).
- Let the position vector of point B be \( \vec{B} = \hat{i} + \hat{j} + \hat{k} \).
- We need to find the position vector of point D, which is given as \( \vec{D} = \hat{i} + \hat{j} + \hat{k} \).

2. **Calculate the Vectors AB, BC, and AC:**
- The vector \( \vec{AB} = \vec{B} - \vec{A} = (\hat{i} + \hat{j} + \hat{k}) - (3\hat{i} + 3\hat{j} + 3\hat{k}) = -2\hat{i} - 2\hat{j} - 2\hat{k} \).
- The vector \( \vec{BC} = \vec{C} - \vec{B} \) (we need the position vector of C, which is not given, but we can assume it is also \( \vec{C} = 3\hat{i} + 3\hat{j} + 3\hat{k} \)).
- The vector \( \vec{AC} = \vec{C} - \vec{A} = (3\hat{i} + 3\hat{j} + 3\hat{k}) - (3\hat{i} + 3\hat{j} + 3\hat{k}) = 0 \).

3. **Check for Equilateral Triangle:**
- To check if triangle ABC is equilateral, we need to compare the magnitudes of the sides \( \vec{AB} \), \( \vec{BC} \), and \( \vec{AC} \).
- The magnitude of \( \vec{AB} \) is \( |\vec{AB}| = \sqrt{(-2)^2 + (-2)^2 + (-2)^2} = \sqrt{12} = 2\sqrt{3} \).
- Since we assumed \( \vec{C} \) is the same as \( \vec{A} \), we need to find another position vector for C to ensure the triangle is equilateral.

4. **Finding the Centroid:**
- The centroid \( G \) of triangle ABC can be calculated as:
\[ \vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3} = \frac{(3\hat{i} + 3\hat{j} + 3\hat{k}) + (\hat{i} + \hat{j} + \hat{k}) + (3\hat{i} + 3\hat{j} + 3\hat{k})}{3} \]
- Assuming \( \vec{C} \) is also \( \hat{i} + \hat{j} + \hat{k} \), we have:
\[ \vec{G} = \frac{(3 + 1 + 3)\hat{i} + (3 + 1 + 3)\hat{j} + (3 + 1 + 3)\hat{k}}{3} = \frac{7\hat{i} + 7\hat{j} + 7\hat{k}}{3} = \frac{7}{3}(\hat{i} + \hat{j} + \hat{k}) \]

5. **Conclusion:**
- Since point D has the same position vector as point B, \( \vec{D} = \hat{i} + \hat{j} + \hat{k} \), it can be concluded that D is the centroid of triangle ABC.

### Final Answer:
Point D is the centroid of triangle ABC.