If the volume of the parallelopiped formed by the vectors `veca, vecb, vecc` as three coterminous edges is 27 units, then the volume of the parallelopiped having `vec(alpha)=veca+2vecb-vecc, vec(beta)=veca-vecb`
and `vec(gamma)=veca-vecb-vecc` as three coterminous edges, is

Options

A. 27
B. 9
C. 81
D. none of these

Answer

To solve the problem, we need to find the volume of the parallelepiped formed by the vectors \(\vec{\alpha}\), \(\vec{\beta}\), and \(\vec{\gamma}\) given that the volume of the parallelepiped formed by the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) is 27 units.

### Step-by-Step Solution:

1. **Identify the given vectors:**
- \(\vec{\alpha} = \vec{a} + 2\vec{b} - \vec{c}\)
- \(\vec{\beta} = \vec{a} - \vec{b}\)
- \(\vec{\gamma} = \vec{a} - \vec{b} - \vec{c}\)

2. **Volume of the parallelepiped:**
The volume \(V\) of a parallelepiped formed by three vectors \(\vec{u}\), \(\vec{v}\), and \(\vec{w}\) can be calculated using the scalar triple product:
\[ V = |\vec{u} \cdot (\vec{v} \times \vec{w})| \]
In our case, we need to find \(V_2 = |\vec{\alpha} \cdot (\vec{\beta} \times \vec{\gamma})|\).

3. **Set up the determinant:**
The volume can also be computed using the determinant of a matrix formed by the coefficients of the vectors:
\[ V_2 = \begin{vmatrix} 1 & 2 & -1 \\ 1 & -1 & 0 \\ 1 & -1 & -1 \end{vmatrix} \]

4. **Calculate the determinant:**
We can expand the determinant:
\[ V_2 = 1 \cdot \begin{vmatrix} -1 & 0 \\ -1 & -1 \end{vmatrix} - 2 \cdot \begin{vmatrix} 1 & 0 \\ 1 & -1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 1 & -1 \\ 1 & -1 \end{vmatrix} \]

Now, calculate each of the 2x2 determinants:
- \(\begin{vmatrix} -1 & 0 \\ -1 & -1 \end{vmatrix} = (-1)(-1) - (0)(-1) = 1\)
- \(\begin{vmatrix} 1 & 0 \\ 1 & -1 \end{vmatrix} = (1)(-1) - (0)(1) = -1\)
- \(\begin{vmatrix} 1 & -1 \\ 1 & -1 \end{vmatrix} = (1)(-1) - (-1)(1) = 0\)

Substitute these values back into the determinant:
\[ V_2 = 1 \cdot 1 - 2 \cdot (-1) + (-1) \cdot 0 = 1 + 2 + 0 = 3 \]

5. **Relate to the original volume:**
Since the volume of the parallelepiped formed by \(\vec{a}, \vec{b}, \vec{c}\) is given as 27, we have:
\[ V_2 = 3 \cdot V_1 = 3 \cdot 27 = 81 \]

### Final Answer:
The volume of the parallelepiped formed by the vectors \(\vec{\alpha}\), \(\vec{\beta}\), and \(\vec{\gamma}\) is **81 cubic units**.