Let `veca, vecb, vecc` be three non-zero non coplanar vectors and `vecp, vecq` and `vecr` be three vectors given by `vecp=veca+vecb-2vecc, vecq=3veca-2vecb+vecc` and `vecr=veca-4vecb+2vecc`
If the volume of the parallelopiped determined by `veca, vecb` and `vecc` is `V_(1)` and that of the parallelopiped determined by `vecp, vecq` and `vecr` is `V_(2)`, then `V_(2):V_(1)=`

Options

A. `3:1`
B. `7:1`
C. `11:1`
D. `15:1`

Answer

To solve the problem, we need to find the ratio of the volumes \( V_2 : V_1 \) of the parallelepipeds formed by the vectors \( \vec{a}, \vec{b}, \vec{c} \) and \( \vec{p}, \vec{q}, \vec{r} \).

### Step-by-Step Solution:

1. **Define the Volume of the Parallelepiped**:
The volume \( V \) of a parallelepiped formed by vectors \( \vec{u}, \vec{v}, \vec{w} \) can be expressed using the scalar triple product:
\[ V = |\vec{u} \cdot (\vec{v} \times \vec{w})| \]

2. **Calculate \( V_1 \)**:
For the vectors \( \vec{a}, \vec{b}, \vec{c} \), the volume \( V_1 \) is given by:
\[ V_1 = |\vec{a} \cdot (\vec{b} \times \vec{c})| \]

3. **Express the Vectors \( \vec{p}, \vec{q}, \vec{r} \)**:
Given:
\[ \vec{p} = \vec{a} + \vec{b} - 2\vec{c} \]
\[ \vec{q} = 3\vec{a} - 2\vec{b} + \vec{c} \]
\[ \vec{r} = \vec{a} - 4\vec{b} + 2\vec{c} \]

4. **Formulate the Determinant for \( V_2 \)**:
The volume \( V_2 \) can be calculated as:
\[ V_2 = |\vec{p} \cdot (\vec{q} \times \vec{r})| \]
To find this, we need to compute the determinant of the matrix formed by \( \vec{p}, \vec{q}, \vec{r} \):
\[ V_2 = \begin{vmatrix} 1 & 1 & -2 \\ 3 & -2 & 1 \\ 1 & -4 & 2 \end{vmatrix} \]

5. **Calculate the Determinant**:
Using the determinant formula:
\[ V_2 = 1 \cdot \begin{vmatrix} -2 & 1 \\ -4 & 2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} - 2 \cdot \begin{vmatrix} 3 & -2 \\ 1 & -4 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
- \( \begin{vmatrix} -2 & 1 \\ -4 & 2 \end{vmatrix} = (-2)(2) - (1)(-4) = -4 + 4 = 0 \)
- \( \begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} = (3)(2) - (1)(1) = 6 - 1 = 5 \)
- \( \begin{vmatrix} 3 & -2 \\ 1 & -4 \end{vmatrix} = (3)(-4) - (-2)(1) = -12 + 2 = -10 \)

Plugging these back into the determinant:
\[ V_2 = 1 \cdot 0 - 1 \cdot 5 - 2 \cdot (-10) = 0 - 5 + 20 = 15 \]

6. **Find the Ratio \( V_2 : V_1 \)**:
Since \( V_1 \) is the volume of the parallelepiped formed by \( \vec{a}, \vec{b}, \vec{c} \) (which we can assume to be 1 for simplicity), we have:
\[ V_1 = 1 \]
Thus, the ratio \( V_2 : V_1 \) is:
\[ V_2 : V_1 = 15 : 1 \]

### Final Answer:
\[ V_2 : V_1 = 15 : 1 \]