Given two vectors `veca= -hati +hatj + 2hatk and vecb =- hati - 2 hatj -hatk`
Find
a. `vecaxxvecb` then use this to find the area of the triangle.
b. The area of the parallelogram
c. The area of a paralleogram whose diagonals are
2 veca and 4vecb

Explanation

To solve the given problem step by step, we will break it down into parts as specified in the question.

### Given Vectors:
- \(\vec{a} = -\hat{i} + \hat{j} + 2\hat{k}\)
- \(\vec{b} = -\hat{i} - 2\hat{j} - \hat{k}\)

### Part (a): Find \(\vec{a} \times \vec{b}\) and the area of the triangle.

1. **Calculate the Cross Product \(\vec{a} \times \vec{b}\)**:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 2 \\ -1 & -2 & -1 \end{vmatrix} \]
Expanding the determinant:
\[ = \hat{i} \begin{vmatrix} 1 & 2 \\ -2 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & 2 \\ -1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 1 \\ -1 & -2 \end{vmatrix} \]
\[ = \hat{i} (1 \cdot -1 - 2 \cdot -2) - \hat{j} (-1 \cdot -1 - 2 \cdot -1) + \hat{k} (-1 \cdot -2 - 1 \cdot -1) \]
\[ = \hat{i} (-1 + 4) - \hat{j} (1 + 2) + \hat{k} (2 + 1) \]
\[ = 3\hat{i} - 3\hat{j} + 3\hat{k} \]
Therefore,
\[ \vec{a} \times \vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k} \]

2. **Find the Magnitude of \(\vec{a} \times \vec{b}\)**:
\[ |\vec{a} \times \vec{b}| = \sqrt{(3)^2 + (-3)^2 + (3)^2} = \sqrt{9 + 9 + 9} = \sqrt{27} = 3\sqrt{3} \]

3. **Calculate the Area of the Triangle**:
The area \(A\) of the triangle formed by vectors \(\vec{a}\) and \(\vec{b}\) is given by:
\[ A = \frac{1}{2} |\vec{a} \times \vec{b}| \]
\[ A = \frac{1}{2} \times 3\sqrt{3} = \frac{3\sqrt{3}}{2} \]

### Part (b): Find the area of the parallelogram.

1. **Area of the Parallelogram**:
The area \(A_p\) of the parallelogram formed by vectors \(\vec{a}\) and \(\vec{b}\) is given by:
\[ A_p = |\vec{a} \times \vec{b}| \]
From the previous calculation, we have:
\[ A_p = 3\sqrt{3} \]

### Part (c): Find the area of a parallelogram whose diagonals are \(2\vec{a}\) and \(4\vec{b}\).

1. **Calculate the Diagonals**:
\[ \text{Diagonal 1} = 2\vec{a} = 2(-\hat{i} + \hat{j} + 2\hat{k}) = -2\hat{i} + 2\hat{j} + 4\hat{k} \]
\[ \text{Diagonal 2} = 4\vec{b} = 4(-\hat{i} - 2\hat{j} - \hat{k}) = -4\hat{i} - 8\hat{j} - 4\hat{k} \]

2. **Calculate the Cross Product of the Diagonals**:
\[ \text{Let } \vec{d_1} = -2\hat{i} + 2\hat{j} + 4\hat{k}, \quad \vec{d_2} = -4\hat{i} - 8\hat{j} - 4\hat{k} \]
\[ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 2 & 4 \\ -4 & -8 & -4 \end{vmatrix} \]
Expanding the determinant:
\[ = \hat{i} \begin{vmatrix} 2 & 4 \\ -8 & -4 \end{vmatrix} - \hat{j} \begin{vmatrix} -2 & 4 \\ -4 & -4 \end{vmatrix} + \hat{k} \begin{vmatrix} -2 & 2 \\ -4 & -8 \end{vmatrix} \]
\[ = \hat{i} (2 \cdot -4 - 4 \cdot -8) - \hat{j} (-2 \cdot -4 - 4 \cdot -4) + \hat{k} (-2 \cdot -8 - 2 \cdot -4) \]
\[ = \hat{i} (-8 + 32) - \hat{j} (8 + 16) + \hat{k} (16 + 8) \]
\[ = 24\hat{i} - 24\hat{j} + 24\hat{k} \]

3. **Find the Magnitude of \(\vec{d_1} \times \vec{d_2}\)**:
\[ |\vec{d_1} \times \vec{d_2}| = \sqrt{(24)^2 + (-24)^2 + (24)^2} = \sqrt{576 + 576 + 576} = \sqrt{1728} = 24\sqrt{3} \]

4. **Calculate the Area of the Parallelogram**:
The area \(A_{p2}\) of the parallelogram formed by the diagonals is given by:
\[ A_{p2} = \frac{1}{2} |\vec{d_1} \times \vec{d_2}| \]
\[ A_{p2} = \frac{1}{2} \times 24\sqrt{3} = 12\sqrt{3} \]

### Final Answers:
- (a) Area of the triangle: \(\frac{3\sqrt{3}}{2}\)
- (b) Area of the parallelogram: \(3\sqrt{3}\)
- (c) Area of the parallelogram with diagonals: \(12\sqrt{3}\)