If the vectors `(hati+hatj+hatk)` and `(3 hati+ 2 hatj)` form two sides of a triangle, then area of the triangle is :

Explanation

To find the area of the triangle formed by the vectors \( \mathbf{A} = \hat{i} + \hat{j} + \hat{k} \) and \( \mathbf{B} = 3\hat{i} + 2\hat{j} \), we can use the formula for the area of a triangle given by two vectors:

\[ \text{Area} = \frac{1}{2} \left| \mathbf{A} \times \mathbf{B} \right| \]

### Step 1: Calculate the cross product \( \mathbf{A} \times \mathbf{B} \)

The vectors are:
- \( \mathbf{A} = \hat{i} + \hat{j} + \hat{k} \)
- \( \mathbf{B} = 3\hat{i} + 2\hat{j} + 0\hat{k} \)

To calculate the cross product, we can use the determinant method:

\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 3 & 2 & 0 \end{vmatrix} \]

### Step 2: Expand the determinant

Using the determinant formula, we can expand it as follows:

\[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} 1 & 1 \\ 2 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 1 \\ 3 & 2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} 1 & 1 \\ 2 & 0 \end{vmatrix} = (1)(0) - (1)(2) = -2 \]

2. For \( -\hat{j} \):
\[ \begin{vmatrix} 1 & 1 \\ 3 & 0 \end{vmatrix} = (1)(0) - (1)(3) = -3 \quad \Rightarrow \quad -(-3) = 3 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 1 & 1 \\ 3 & 2 \end{vmatrix} = (1)(2) - (1)(3) = 2 - 3 = -1 \]

Putting it all together:

\[ \mathbf{A} \times \mathbf{B} = -2\hat{i} + 3\hat{j} - 1\hat{k} \]

### Step 3: Find the magnitude of the cross product

Now we find the magnitude of the vector \( -2\hat{i} + 3\hat{j} - 1\hat{k} \):

\[ \left| \mathbf{A} \times \mathbf{B} \right| = \sqrt{(-2)^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14} \]

### Step 4: Calculate the area of the triangle

Now we can find the area:

\[ \text{Area} = \frac{1}{2} \left| \mathbf{A} \times \mathbf{B} \right| = \frac{1}{2} \sqrt{14} \]

### Final Answer

Thus, the area of the triangle is:

\[ \text{Area} = \frac{\sqrt{14}}{2} \text{ square units} \]