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 need to find the areas of a triangle and two parallelograms based on the vectors \(\vec{A}\) and \(\vec{B}\).

Given:
\[ \vec{A} = 3\hat{i} + \hat{j} + \hat{k} \]
\[ \vec{B} = \hat{i} - \hat{j} - \hat{k} \]

### Step 1: Calculate \(\vec{A} \times \vec{B}\)

To find the cross product \(\vec{A} \times \vec{B}\), we can use the determinant method:

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

Calculating the determinant:

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

Calculating each of the 2x2 determinants:

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

2. For \(-\hat{j}\):
\[ \begin{vmatrix} 3 & 1 \\ 1 & -1 \end{vmatrix} = (3)(-1) - (1)(1) = -3 - 1 = -4 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} 3 & 1 \\ 1 & -1 \end{vmatrix} = (3)(-1) - (1)(1) = -3 - 1 = -4 \]

Putting it all together:
\[ \vec{A} \times \vec{B} = 0\hat{i} + 4\hat{j} - 4\hat{k} = 4\hat{j} - 4\hat{k} \]

### Step 2: Calculate the magnitude of \(\vec{A} \times \vec{B}\)

The magnitude is given by:
\[ |\vec{A} \times \vec{B}| = \sqrt{(0)^2 + (4)^2 + (-4)^2} = \sqrt{0 + 16 + 16} = \sqrt{32} = 4\sqrt{2} \]

### Step 3: Calculate the areas

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

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

c. **Area of the parallelogram whose diagonals are \(\vec{A}\) and \(\vec{B}\)**:
The area is given by:
\[ \text{Area}_{\text{parallelogram (diagonals)}} = \frac{1}{2} |\vec{A} \times \vec{B}| = \frac{1}{2} (4\sqrt{2}) = 2\sqrt{2} \]

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