If the position vectors of the vertices A,B and C of a `DeltaABC` are `7hatj+10k,-hati+6hatj+6hatk and -4hati+9hatj+6hatk`, respectively, the triangle is

Options

A. A. equilateral
B. B. isosceles
C. C. scalene
D. D. right angled and isosceles also

Explanation

To determine the type of triangle formed by the vertices A, B, and C with given position vectors, we will follow these steps:

### Step 1: Define the Position Vectors
Let:
- Position vector of A, **A** = \( 0\hat{i} + 7\hat{j} + 10\hat{k} \)
- Position vector of B, **B** = \( -1\hat{i} + 6\hat{j} + 6\hat{k} \)
- Position vector of C, **C** = \( -4\hat{i} + 9\hat{j} + 6\hat{k} \)

### Step 2: Calculate the Vectors AB, BC, and AC
1. **Vector AB**:
\[ \vec{AB} = \vec{B} - \vec{A} = (-1\hat{i} + 6\hat{j} + 6\hat{k}) - (0\hat{i} + 7\hat{j} + 10\hat{k}) \]
\[ = -1\hat{i} + (6 - 7)\hat{j} + (6 - 10)\hat{k} = -1\hat{i} - 1\hat{j} - 4\hat{k} \]

2. **Vector BC**:
\[ \vec{BC} = \vec{C} - \vec{B} = (-4\hat{i} + 9\hat{j} + 6\hat{k}) - (-1\hat{i} + 6\hat{j} + 6\hat{k}) \]
\[ = (-4 + 1)\hat{i} + (9 - 6)\hat{j} + (6 - 6)\hat{k} = -3\hat{i} + 3\hat{j} + 0\hat{k} \]

3. **Vector AC**:
\[ \vec{AC} = \vec{C} - \vec{A} = (-4\hat{i} + 9\hat{j} + 6\hat{k}) - (0\hat{i} + 7\hat{j} + 10\hat{k}) \]
\[ = -4\hat{i} + (9 - 7)\hat{j} + (6 - 10)\hat{k} = -4\hat{i} + 2\hat{j} - 4\hat{k} \]

### Step 3: Calculate the Magnitudes of the Vectors
1. **Magnitude of AB**:
\[ |\vec{AB}| = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} = \sqrt{1 + 1 + 16} = \sqrt{18} \]

2. **Magnitude of BC**:
\[ |\vec{BC}| = \sqrt{(-3)^2 + (3)^2 + (0)^2} = \sqrt{9 + 9} = \sqrt{18} \]

3. **Magnitude of AC**:
\[ |\vec{AC}| = \sqrt{(-4)^2 + (2)^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6 \]

### Step 4: Determine the Type of Triangle
- We have:
- \( |\vec{AB}| = \sqrt{18} \)
- \( |\vec{BC}| = \sqrt{18} \)
- \( |\vec{AC}| = 6 \)

Since \( |\vec{AB}| = |\vec{BC}| \), the triangle is **isosceles**.

### Conclusion
The triangle formed by the vertices A, B, and C is an **isosceles triangle**.

---