Let the vectors $$\vec{a}, \vec{b}, \vec{c}$$ represent three coterminous edges of a parallelopiped of volume $$V$$. Then the volume of the parallelopiped, whose coterminous edges are represented by $$\vec{a}, \vec{b}+\vec{c}$$ and $$\vec{a}+2\vec{b}+3\vec{c}$$ is equal to

Given that $$\vec{a}, \vec{b}, \vec{c}$$ are coterminous edges of a parallelepiped with volume $$V = [\vec{a} \; \vec{b} \; \vec{c}]$$.

We need to find the volume of the parallelepiped with edges $$\vec{a}$$, $$\vec{b} + \vec{c}$$, and $$\vec{a} + 2\vec{b} + 3\vec{c}$$.

Write the volume as a scalar triple product.

$$V' = [\vec{a}, \; \vec{b}+\vec{c}, \; \vec{a}+2\vec{b}+3\vec{c}]$$

$$= \vec{a} \cdot \left[(\vec{b}+\vec{c}) \times (\vec{a}+2\vec{b}+3\vec{c})\right]$$

Expand the cross product.

$$(\vec{b}+\vec{c}) \times (\vec{a}+2\vec{b}+3\vec{c})$$

$$= \vec{b} \times \vec{a} + 2(\vec{b} \times \vec{b}) + 3(\vec{b} \times \vec{c}) + \vec{c} \times \vec{a} + 2(\vec{c} \times \vec{b}) + 3(\vec{c} \times \vec{c})$$

Since $$\vec{b} \times \vec{b} = \vec{0}$$ and $$\vec{c} \times \vec{c} = \vec{0}$$, and $$\vec{c} \times \vec{b} = -\vec{b} \times \vec{c}$$:

$$= \vec{b} \times \vec{a} + 3(\vec{b} \times \vec{c}) + \vec{c} \times \vec{a} - 2(\vec{b} \times \vec{c})$$

$$= \vec{b} \times \vec{a} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$$

Take the dot product with $$\vec{a}$$.

$$V' = \vec{a} \cdot (\vec{b} \times \vec{a}) + \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{a} \cdot (\vec{c} \times \vec{a})$$

Since $$\vec{a} \cdot (\vec{b} \times \vec{a}) = 0$$ (coplanar vectors) and $$\vec{a} \cdot (\vec{c} \times \vec{a}) = 0$$ (coplanar vectors):

$$V' = \vec{a} \cdot (\vec{b} \times \vec{c}) = [\vec{a} \; \vec{b} \; \vec{c}] = V$$

Therefore, the volume of the new parallelepiped is $$V$$, which is Option C.