The volume of a tetrahedron formed by the coterminous edges ` vec a , vec b ,a n d vec c` is 3. Then the volume of the parallelepiped formed by the coterminous edges ` vec a+ vec b , vec b+ vec ca n d vec c+ vec a` is `6` b. `18` c. `36` d. 9

Explanation

To solve the problem, we need to find the volume of the parallelepiped formed by the vectors \( \vec{a} + \vec{b} \), \( \vec{b} + \vec{c} \), and \( \vec{c} + \vec{a} \), given that the volume of the tetrahedron formed by the vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) is 3.

### Step-by-Step Solution:

1. **Volume of the Tetrahedron**:
The volume \( V \) of a tetrahedron formed by vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) is given by the formula:
\[
V = \frac{1}{6} |\vec{a} \cdot (\vec{b} \times \vec{c})|
...