The points A, B and C with position vectors `a=3i-4j-4k,b=2i-j+kandc=i-3j-5k`, respectively are

Options

A. Collinear
B. Vertices of an equilateral triangle
C. Vertices of a right angled triangle
D. none of these

Answer

To determine the relationship between the points A, B, and C with the given position vectors, we will follow these steps:

### Step 1: Write down the position vectors
The position vectors of the points A, B, and C are given as:
- \( \vec{A} = 3\hat{i} - 4\hat{j} - 4\hat{k} \)
- \( \vec{B} = 2\hat{i} - \hat{j} + \hat{k} \)
- \( \vec{C} = \hat{i} - 3\hat{j} - 5\hat{k} \)

### Step 2: Find the vectors AB and AC
To check if the points are collinear, we need to find the vectors \( \vec{AB} \) and \( \vec{AC} \).

1. **Calculate \( \vec{AB} \)**:
\[ \vec{AB} = \vec{B} - \vec{A} = (2\hat{i} - \hat{j} + \hat{k}) - (3\hat{i} - 4\hat{j} - 4\hat{k}) \]
\[ = (2 - 3)\hat{i} + (-1 + 4)\hat{j} + (1 + 4)\hat{k} = -\hat{i} + 3\hat{j} + 5\hat{k} \]

2. **Calculate \( \vec{AC} \)**:
\[ \vec{AC} = \vec{C} - \vec{A} = (\hat{i} - 3\hat{j} - 5\hat{k}) - (3\hat{i} - 4\hat{j} - 4\hat{k}) \]
\[ = (1 - 3)\hat{i} + (-3 + 4)\hat{j} + (-5 + 4)\hat{k} = -2\hat{i} + 1\hat{j} - 1\hat{k} \]

### Step 3: Check for collinearity
To check if the points A, B, and C are collinear, we can see if the vectors \( \vec{AB} \) and \( \vec{AC} \) are scalar multiples of each other. This can be done by checking if the cross product \( \vec{AB} \times \vec{AC} \) equals zero.

1. **Calculate the cross product \( \vec{AB} \times \vec{AC} \)**:
\[ \vec{AB} = -\hat{i} + 3\hat{j} + 5\hat{k} \]
\[ \vec{AC} = -2\hat{i} + 1\hat{j} - 1\hat{k} \]
Using the determinant method:
\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & 5 \\ -2 & 1 & -1 \end{vmatrix} \]
Expanding this determinant:
\[ = \hat{i} \begin{vmatrix} 3 & 5 \\ 1 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & 5 \\ -2 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 3 \\ -2 & 1 \end{vmatrix} \]
Calculating the minors:
\[ = \hat{i} (3 \cdot -1 - 5 \cdot 1) - \hat{j} (-1 \cdot -1 - 5 \cdot -2) + \hat{k} (-1 \cdot 1 - 3 \cdot -2) \]
\[ = \hat{i} (-3 - 5) - \hat{j} (1 + 10) + \hat{k} (-1 + 6) \]
\[ = -8\hat{i} - 11\hat{j} + 5\hat{k} \]

Since the cross product is not equal to zero, the vectors \( \vec{AB} \) and \( \vec{AC} \) are not collinear.

### Conclusion
The points A, B, and C are not collinear. Therefore, they do not form the vertices of an equilateral triangle or a right angle triangle.