If the vectors `6hati-2hatj+3hatkk,2hati+3hatj-6hatk and 3hati+6hatj-2hatk` form a triangle, then it is

Options

A. right angled
B. obtuse angled
C. equilateral
D. isosceles

Explanation

To determine the type of triangle formed by the given vectors \( \vec{A} = 6\hat{i} - 2\hat{j} + 3\hat{k} \), \( \vec{B} = 2\hat{i} + 3\hat{j} - 6\hat{k} \), and \( \vec{C} = 3\hat{i} + 6\hat{j} - 2\hat{k} \), we will follow these steps:

### Step 1: Define the position vectors
Let:
- Position vector of point A: \( \vec{A} = 6\hat{i} - 2\hat{j} + 3\hat{k} \)
- Position vector of point B: \( \vec{B} = 2\hat{i} + 3\hat{j} - 6\hat{k} \)
- Position vector of point C: \( \vec{C} = 3\hat{i} + 6\hat{j} - 2\hat{k} \)

### Step 2: Calculate the vector \( \vec{AB} \)
The vector \( \vec{AB} \) is given by:
\[ \vec{AB} = \vec{B} - \vec{A} = (2\hat{i} + 3\hat{j} - 6\hat{k}) - (6\hat{i} - 2\hat{j} + 3\hat{k}) \]
Calculating this gives:
\[ \vec{AB} = (2 - 6)\hat{i} + (3 + 2)\hat{j} + (-6 - 3)\hat{k} = -4\hat{i} + 5\hat{j} - 9\hat{k} \]

### Step 3: Calculate the magnitude of \( \vec{AB} \)
The magnitude of \( \vec{AB} \) is:
\[ |\vec{AB}| = \sqrt{(-4)^2 + 5^2 + (-9)^2} = \sqrt{16 + 25 + 81} = \sqrt{122} \]

### Step 4: Calculate the vector \( \vec{BC} \)
The vector \( \vec{BC} \) is given by:
\[ \vec{BC} = \vec{C} - \vec{B} = (3\hat{i} + 6\hat{j} - 2\hat{k}) - (2\hat{i} + 3\hat{j} - 6\hat{k}) \]
Calculating this gives:
\[ \vec{BC} = (3 - 2)\hat{i} + (6 - 3)\hat{j} + (-2 + 6)\hat{k} = 1\hat{i} + 3\hat{j} + 4\hat{k} \]

### Step 5: Calculate the magnitude of \( \vec{BC} \)
The magnitude of \( \vec{BC} \) is:
\[ |\vec{BC}| = \sqrt{1^2 + 3^2 + 4^2} = \sqrt{1 + 9 + 16} = \sqrt{26} \]

### Step 6: Calculate the vector \( \vec{AC} \)
The vector \( \vec{AC} \) is given by:
\[ \vec{AC} = \vec{C} - \vec{A} = (3\hat{i} + 6\hat{j} - 2\hat{k}) - (6\hat{i} - 2\hat{j} + 3\hat{k}) \]
Calculating this gives:
\[ \vec{AC} = (3 - 6)\hat{i} + (6 + 2)\hat{j} + (-2 - 3)\hat{k} = -3\hat{i} + 8\hat{j} - 5\hat{k} \]

### Step 7: Calculate the magnitude of \( \vec{AC} \)
The magnitude of \( \vec{AC} \) is:
\[ |\vec{AC}| = \sqrt{(-3)^2 + 8^2 + (-5)^2} = \sqrt{9 + 64 + 25} = \sqrt{98} \]

### Step 8: Compare the squares of the magnitudes
Now we will compare \( |\vec{AB}|^2 \), \( |\vec{BC}|^2 \), and \( |\vec{AC}|^2 \):
- \( |\vec{AB}|^2 = 122 \)
- \( |\vec{BC}|^2 = 26 \)
- \( |\vec{AC}|^2 = 98 \)

We check if \( |\vec{AC}|^2 < |\vec{AB}|^2 + |\vec{BC}|^2 \):
\[ 98 < 122 + 26 \Rightarrow 98 < 148 \]
This condition holds true.

### Conclusion
Since \( |\vec{AC}|^2 < |\vec{AB}|^2 + |\vec{BC}|^2 \), triangle \( ABC \) is an **obtuse triangle**.

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