Given two vectors `vec(A)=3hat(i)+hat(j)+hat(k)` and `vec(B)=hat(i)-hat(j)-hat(k)`.Find the
a.Area of the triangle whose two adjacent sides are represented by the vector `vec(A)` and `vec(B)`
b.Area of the parallelogram whose two adjacent sides are represented by the vector `vec(A)` and `vec(B)`
c. Area of the parallelogram whose diagnoals are represented by the vector `vec(A)` and `vec(B)`

Explanation

To solve the given problem, we will follow these steps:

### Given:
- Vector \( \vec{A} = 3\hat{i} + \hat{j} + \hat{k} \)
- Vector \( \vec{B} = \hat{i} - \hat{j} - \hat{k} \)

### Part (a): Area of the triangle whose two adjacent sides are represented by the vectors \( \vec{A} \) and \( \vec{B} \)

1. **Calculate the cross product \( \vec{A} \times \vec{B} \)**:
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 1 \\ 1 & -1 & -1 \end{vmatrix} \]

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

Calculating the determinants:
\[ = \hat{i} (1 \cdot -1 - 1 \cdot -1) - \hat{j} (3 \cdot -1 - 1 \cdot 1) + \hat{k} (3 \cdot -1 - 1 \cdot 1) \]
\[ = \hat{i} (0) - \hat{j} (-4) + \hat{k} (-4) \]
\[ = 4\hat{j} - 4\hat{k} \]

2. **Calculate the magnitude of the cross product**:
\[ |\vec{A} \times \vec{B}| = \sqrt{(0)^2 + (4)^2 + (-4)^2} = \sqrt{0 + 16 + 16} = \sqrt{32} = 4\sqrt{2} \]

3. **Area of the triangle**:
\[ \text{Area} = \frac{1}{2} |\vec{A} \times \vec{B}| = \frac{1}{2} (4\sqrt{2}) = 2\sqrt{2} \]

### Part (b): Area of the parallelogram whose two adjacent sides are represented by the vectors \( \vec{A} \) and \( \vec{B} \)

1. **Area of the parallelogram**:
\[ \text{Area} = |\vec{A} \times \vec{B}| = 4\sqrt{2} \]

### Part (c): Area of the parallelogram whose diagonals are represented by the vectors \( \vec{A} \) and \( \vec{B} \)

1. **Area of the parallelogram with diagonals**:
\[ \text{Area} = \frac{1}{2} |\vec{A} \times \vec{B}| \]
\[ = \frac{1}{2} (4\sqrt{2}) = 2\sqrt{2} \]

### Summary of Answers:
- a. Area of the triangle: \( 2\sqrt{2} \)
- b. Area of the parallelogram: \( 4\sqrt{2} \)
- c. Area of the parallelogram with diagonals: \( 2\sqrt{2} \)