Vertices of a triangle are given by `hat(i) + 3hat(j) + 2hat(k), 2hat(i) - hat(j) + hat(k) and -hat(i) + 2hat(j) + 3hat(k)`, then the area of triangle is `sqrt((107)/(2n))` (in units). Find the value of n=

Answer

To find the area of the triangle formed by the given vertices, we will follow these steps:

### Step 1: Identify the vertices
The vertices of the triangle are given as:
- \( A = \hat{i} + 3\hat{j} + 2\hat{k} \)
- \( B = 2\hat{i} - \hat{j} + \hat{k} \)
- \( C = -\hat{i} + 2\hat{j} + 3\hat{k} \)

### Step 2: Find the position vectors
Let:
- \( \vec{OA} = \hat{i} + 3\hat{j} + 2\hat{k} \)
- \( \vec{OB} = 2\hat{i} - \hat{j} + \hat{k} \)
- \( \vec{OC} = -\hat{i} + 2\hat{j} + 3\hat{k} \)

### Step 3: Calculate the vectors for two sides of the triangle
We will calculate the vectors \( \vec{BA} \) and \( \vec{BC} \):
- \( \vec{BA} = \vec{OA} - \vec{OB} = (\hat{i} + 3\hat{j} + 2\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) \)
\[ = \hat{i} - 2\hat{i} + 3\hat{j} + \hat{j} + 2\hat{k} - \hat{k} = -\hat{i} + 4\hat{j} + \hat{k} \]
- \( \vec{BC} = \vec{OC} - \vec{OB} = (-\hat{i} + 2\hat{j} + 3\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) \)
\[ = -\hat{i} - 2\hat{i} + 2\hat{j} + \hat{j} + 3\hat{k} - \hat{k} = -3\hat{i} + 3\hat{j} + 2\hat{k} \]

### Step 4: Calculate the cross product \( \vec{BA} \times \vec{BC} \)
We will use the determinant method to find the cross product:
\[ \vec{BA} \times \vec{BC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 4 & 1 \\ -3 & 3 & 2 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 4 & 1 \\ 3 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & 1 \\ -3 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 4 \\ -3 & 3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 4 & 1 \\ 3 & 2 \end{vmatrix} = (4)(2) - (1)(3) = 8 - 3 = 5 \)
2. \( \begin{vmatrix} -1 & 1 \\ -3 & 2 \end{vmatrix} = (-1)(2) - (1)(-3) = -2 + 3 = 1 \)
3. \( \begin{vmatrix} -1 & 4 \\ -3 & 3 \end{vmatrix} = (-1)(3) - (4)(-3) = -3 + 12 = 9 \)

Putting it all together:
\[ \vec{BA} \times \vec{BC} = 5\hat{i} - 1\hat{j} + 9\hat{k} \]

### Step 5: Calculate the magnitude of the cross product
\[ |\vec{BA} \times \vec{BC}| = \sqrt{5^2 + (-1)^2 + 9^2} = \sqrt{25 + 1 + 81} = \sqrt{107} \]

### Step 6: Calculate the area of the triangle
The area \( A \) of the triangle is given by:
\[ A = \frac{1}{2} |\vec{BA} \times \vec{BC}| = \frac{1}{2} \sqrt{107} \]

### Step 7: Compare with the given area
The problem states that the area can also be expressed as:
\[ A = \sqrt{\frac{107}{2n}} \]
Setting the two expressions for area equal:
\[ \frac{1}{2} \sqrt{107} = \sqrt{\frac{107}{2n}} \]

### Step 8: Solve for \( n \)
Squaring both sides:
\[ \left(\frac{1}{2}\right)^2 \cdot 107 = \frac{107}{2n} \]
\[ \frac{107}{4} = \frac{107}{2n} \]
Cross-multiplying gives:
\[ 107 \cdot 2n = 4 \cdot 107 \]
Dividing both sides by \( 107 \):
\[ 2n = 4 \implies n = 2 \]

### Final Answer
The value of \( n \) is \( 2 \).

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