Find the volume of the parallelepiped whose coterminous edges are represented by the vectors
(i) `vec(a)=hat(i)+hat(j)+hat(k), vec(b)=hat(i)-hat(j)+hat(k), vec(c)=hat(i)+2hat(j)-hat(k)`
(ii) `vec(a)=-3hat(i)+7hat(j)+5hat(k), vec(b)=-5hat(i)+7hat(j)-3hat(k), vec(c)= 7 hat(i)-5hat(j)-3hat(k)`
(iii)`vec(a)=hat(i)-2hat(j)+3hat(k), vec(b)=2hat(i)+hat(j)-hat(k), vec(c)=hat(j)+hat(k)`
(iv) `vec(a)=6hat(i), vec(b)=2hat(j), vec(c)=5hat(k)`

Answer

To find the volume of the parallelepiped formed by the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), we can use the formula:

\[ \text{Volume} = |\vec{a} \cdot (\vec{b} \times \vec{c})| \]

This can also be computed using the determinant of a 3x3 matrix formed by the vectors. Let's solve each part step by step.

### Part (i)
Given vectors:
\[ \vec{a} = \hat{i} + \hat{j} + \hat{k}, \quad \vec{b} = \hat{i} - \hat{j} + \hat{k}, \quad \vec{c} = \hat{i} + 2\hat{j} - \hat{k} \]

1. **Set up the determinant:**
\[ V = \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix} \]

2. **Calculate the determinant:**
\[ V = 1 \begin{vmatrix} -1 & 1 \\ 2 & -1 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} + 1 \begin{vmatrix} 1 & -1 \\ 1 & 2 \end{vmatrix} \]

- First determinant:
\[ (-1)(-1) - (1)(2) = 1 - 2 = -1 \]
- Second determinant:
\[ (1)(-1) - (1)(1) = -1 - 1 = -2 \]
- Third determinant:
\[ (1)(2) - (-1)(1) = 2 + 1 = 3 \]

Putting it all together:
\[ V = 1(-1) - 1(-2) + 1(3) = -1 + 2 + 3 = 4 \]

3. **Final volume:**
\[ \text{Volume} = |4| = 4 \text{ cubic units} \]

### Part (ii)
Given vectors:
\[ \vec{a} = -3\hat{i} + 7\hat{j} + 5\hat{k}, \quad \vec{b} = -5\hat{i} + 7\hat{j} - 3\hat{k}, \quad \vec{c} = 7\hat{i} - 5\hat{j} - 3\hat{k} \]

1. **Set up the determinant:**
\[ V = \begin{vmatrix} -3 & 7 & 5 \\ -5 & 7 & -3 \\ 7 & -5 & -3 \end{vmatrix} \]

2. **Calculate the determinant:**
Using the same method as above, we compute the determinant step by step.

After calculating, we find:
\[ V = 264 \]

3. **Final volume:**
\[ \text{Volume} = |264| = 264 \text{ cubic units} \]

### Part (iii)
Given vectors:
\[ \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}, \quad \vec{b} = 2\hat{i} + \hat{j} - \hat{k}, \quad \vec{c} = \hat{j} + \hat{k} \]

1. **Set up the determinant:**
\[ V = \begin{vmatrix} 1 & -2 & 3 \\ 2 & 1 & -1 \\ 0 & 1 & 1 \end{vmatrix} \]

2. **Calculate the determinant:**
After computing, we find:
\[ V = 12 \]

3. **Final volume:**
\[ \text{Volume} = |12| = 12 \text{ cubic units} \]

### Part (iv)
Given vectors:
\[ \vec{a} = 6\hat{i}, \quad \vec{b} = 2\hat{j}, \quad \vec{c} = 5\hat{k} \]

1. **Set up the determinant:**
\[ V = \begin{vmatrix} 6 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 5 \end{vmatrix} \]

2. **Calculate the determinant:**
\[ V = 6 \cdot 2 \cdot 5 = 60 \]

3. **Final volume:**
\[ \text{Volume} = |60| = 60 \text{ cubic units} \]

### Summary of Volumes:
1. Part (i): 4 cubic units
2. Part (ii): 264 cubic units
3. Part (iii): 12 cubic units
4. Part (iv): 60 cubic units