Statement 1: Let `veca, vecb, vecc` be three coterminous edges of a parallelopiped of volume `V`. Let `V_(1)` be the volume of the parallelopiped whose three coterminous edges are the diagonals of three adjacent faces of the given parallelopiped. Then `V_(1)=2V`.
Statement 2: For any three vectors, `vecp, vecq, vecr`
`[(vecp+vecq, vecq+vecr,vecr+vecp)]=2[(vecp,vecq,vecr)]`

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we will analyze both statements step by step.

### Step 1: Understanding the Volume of the Parallelepiped
Let \(\vec{a}, \vec{b}, \vec{c}\) be the three coterminous edges of a parallelepiped. The volume \(V\) of the parallelepiped formed by these vectors is given by the scalar triple product:
\[ V = |\vec{a} \cdot (\vec{b} \times \vec{c})| \]

### Step 2: Defining the New Volume \(V_1\)
Now, we need to find the volume \(V_1\) of the parallelepiped whose edges are the diagonals of the three adjacent faces of the original parallelepiped. The diagonals of the faces can be expressed as:
- Diagonal of face formed by \(\vec{a}\) and \(\vec{b}\): \(\vec{a} + \vec{b}\)
- Diagonal of face formed by \(\vec{b}\) and \(\vec{c}\): \(\vec{b} + \vec{c}\)
- Diagonal of face formed by \(\vec{c}\) and \(\vec{a}\): \(\vec{c} + \vec{a}\)

Thus, the edges of the new parallelepiped are:
\[ \vec{d_1} = \vec{a} + \vec{b}, \quad \vec{d_2} = \vec{b} + \vec{c}, \quad \vec{d_3} = \vec{c} + \vec{a} \]

### Step 3: Finding Volume \(V_1\)
The volume \(V_1\) can be calculated using the scalar triple product:
\[ V_1 = |(\vec{d_1} \cdot (\vec{d_2} \times \vec{d_3}))| \]

Substituting the expressions for \(\vec{d_1}, \vec{d_2}, \vec{d_3}\):
\[ V_1 = |((\vec{a} + \vec{b}) \cdot ((\vec{b} + \vec{c}) \times (\vec{c} + \vec{a})))| \]

### Step 4: Simplifying the Cross Product
To simplify \((\vec{b} + \vec{c}) \times (\vec{c} + \vec{a})\), we can use the distributive property of the cross product:
\[ (\vec{b} + \vec{c}) \times (\vec{c} + \vec{a}) = \vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{c} + \vec{c} \times \vec{a} \]
Since \(\vec{c} \times \vec{c} = \vec{0}\), we have:
\[ = \vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{a} \]

### Step 5: Final Calculation
Now substituting back into the volume expression:
\[ V_1 = |(\vec{a} + \vec{b}) \cdot (\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{a})| \]
This can be simplified further, and after some calculations, we find:
\[ V_1 = 2V \]

### Conclusion for Statement 1
Thus, Statement 1 is true: \(V_1 = 2V\).

### Step 6: Analyzing Statement 2
For Statement 2, we need to show:
\[ [(\vec{p} + \vec{q}, \vec{q} + \vec{r}, \vec{r} + \vec{p})] = 2[(\vec{p}, \vec{q}, \vec{r})] \]

### Step 7: Setting Up the Determinant
Let:
\[ \vec{u} = \vec{p} + \vec{q}, \quad \vec{v} = \vec{q} + \vec{r}, \quad \vec{w} = \vec{r} + \vec{p} \]
We can express the determinant:
\[ \begin{vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix} \]

### Step 8: Evaluating the Determinant
Calculating the determinant gives us:
\[ = 2 \]
Thus,
\[ [(\vec{u}, \vec{v}, \vec{w})] = 2[(\vec{p}, \vec{q}, \vec{r})] \]

### Conclusion for Statement 2
Therefore, Statement 2 is also true.

### Final Answer
Both statements are true:
1. \(V_1 = 2V\)
2. \([( \vec{p} + \vec{q}, \vec{q} + \vec{r}, \vec{r} + \vec{p})] = 2[(\vec{p}, \vec{q}, \vec{r})]\)

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