Let `a=2i-3j+4k, b=i+2j-2k " and "c=3i-j+k`. Let V be, the volume (in cubic unit) of the parallelopiped having `a+b+c, a-b+c" and "a+b-c` as coterminus edges, then V = ________

Answer

To find the volume \( V \) of the parallelepiped formed by the vectors \( \mathbf{a} + \mathbf{b} + \mathbf{c} \), \( \mathbf{a} - \mathbf{b} + \mathbf{c} \), and \( \mathbf{a} + \mathbf{b} - \mathbf{c} \), we will follow these steps:

### Step 1: Calculate the vectors

1. **Calculate \( \mathbf{a} + \mathbf{b} + \mathbf{c} \)**:
\[ \mathbf{a} = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}, \quad \mathbf{b} = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k}, \quad \mathbf{c} = 3\mathbf{i} - \mathbf{j} + \mathbf{k} \]
\[ \mathbf{a} + \mathbf{b} + \mathbf{c} = (2 + 1 + 3)\mathbf{i} + (-3 + 2 - 1)\mathbf{j} + (4 - 2 + 1)\mathbf{k} \]
\[ = 6\mathbf{i} - 2\mathbf{j} + 3\mathbf{k} \]

2. **Calculate \( \mathbf{a} - \mathbf{b} + \mathbf{c} \)**:
\[ \mathbf{a} - \mathbf{b} + \mathbf{c} = (2 - 1 + 3)\mathbf{i} + (-3 - 2 - 1)\mathbf{j} + (4 - (-2) + 1)\mathbf{k} \]
\[ = 4\mathbf{i} - 6\mathbf{j} + 7\mathbf{k} \]

3. **Calculate \( \mathbf{a} + \mathbf{b} - \mathbf{c} \)**:
\[ \mathbf{a} + \mathbf{b} - \mathbf{c} = (2 + 1 - 3)\mathbf{i} + (-3 + 2 + 1)\mathbf{j} + (4 - (-1))\mathbf{k} \]
\[ = 0\mathbf{i} + 0\mathbf{j} + 5\mathbf{k} \]

### Step 2: Form the matrix for the scalar triple product

The volume \( V \) of the parallelepiped can be calculated using the scalar triple product:
\[ V = |\mathbf{a} + \mathbf{b} + \mathbf{c} \cdot (\mathbf{a} - \mathbf{b} + \mathbf{c}) \times (\mathbf{a} + \mathbf{b} - \mathbf{c})| \]
We can represent the vectors as a matrix:
\[ \begin{vmatrix} 6 & -2 & 3 \\ 4 & -6 & 7 \\ 0 & 0 & 5 \end{vmatrix} \]

### Step 3: Calculate the determinant

To find the determinant, we can expand it:
\[ = 6 \begin{vmatrix} -6 & 7 \\ 0 & 5 \end{vmatrix} - (-2) \begin{vmatrix} 4 & 7 \\ 0 & 5 \end{vmatrix} + 3 \begin{vmatrix} 4 & -6 \\ 0 & 0 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} -6 & 7 \\ 0 & 5 \end{vmatrix} = (-6)(5) - (7)(0) = -30 \)
2. \( \begin{vmatrix} 4 & 7 \\ 0 & 5 \end{vmatrix} = (4)(5) - (7)(0) = 20 \)
3. \( \begin{vmatrix} 4 & -6 \\ 0 & 0 \end{vmatrix} = 0 \)

Now substituting back:
\[ = 6(-30) + 2(20) + 3(0) = -180 + 40 + 0 = -140 \]

### Step 4: Find the volume

The volume is the absolute value of the determinant:
\[ V = | -140 | = 140 \]

### Final Answer:
Thus, the volume \( V \) of the parallelepiped is \( 140 \) cubic units.