If `veca, vecb, vecc` are non-null non coplanar vectors, then
`[(veca-2vecb+vecc, vecb-2vecc+veca, vecc-2veca+vecb)]=`

Options

A. `[(veca, vecb, vecc)]`
B. 3`[(veca, vecb, vecc)]`
C. `0`
D. `12[(veca, vecb, vecc)]`

Answer

To solve the problem, we need to evaluate the scalar triple product of the vectors \( \vec{A} = \vec{a} - 2\vec{b} + \vec{c} \), \( \vec{B} = \vec{b} - 2\vec{c} + \vec{a} \), and \( \vec{C} = \vec{c} - 2\vec{a} + \vec{b} \).

### Step 1: Define the vectors
Let:
\[ \vec{A} = \vec{a} - 2\vec{b} + \vec{c} \]
\[ \vec{B} = \vec{b} - 2\vec{c} + \vec{a} \]
\[ \vec{C} = \vec{c} - 2\vec{a} + \vec{b} \]

### Step 2: Calculate the scalar triple product
We need to find the scalar triple product \( [\vec{A}, \vec{B}, \vec{C}] \), which can be expressed as:
\[ [\vec{A}, \vec{B}, \vec{C}] = \vec{A} \cdot (\vec{B} \times \vec{C}) \]

### Step 3: Compute \( \vec{B} \times \vec{C} \)
Using the properties of the cross product:
\[ \vec{B} \times \vec{C} = (\vec{b} - 2\vec{c} + \vec{a}) \times (\vec{c} - 2\vec{a} + \vec{b}) \]

Expanding this using the distributive property of the cross product:
\[ = \vec{b} \times \vec{c} - 2\vec{b} \times \vec{a} + \vec{b} \times \vec{b} - 2\vec{c} \times \vec{c} + 2\vec{c} \times \vec{a} + \vec{a} \times \vec{b} \]

Since \( \vec{b} \times \vec{b} = 0 \) and \( \vec{c} \times \vec{c} = 0 \):
\[ = \vec{b} \times \vec{c} - 2\vec{b} \times \vec{a} + 2\vec{c} \times \vec{a} + \vec{a} \times \vec{b} \]

### Step 4: Combine like terms
Rearranging the terms gives:
\[ = \vec{b} \times \vec{c} + \vec{a} \times \vec{b} - 2\vec{b} \times \vec{a} + 2\vec{c} \times \vec{a} \]
Using the property \( \vec{u} \times \vec{v} = -\vec{v} \times \vec{u} \):
\[ = \vec{b} \times \vec{c} + \vec{a} \times \vec{b} + 2\vec{a} \times \vec{c} \]

### Step 5: Substitute back into the scalar triple product
Now substitute \( \vec{B} \times \vec{C} \) back into the scalar triple product:
\[ [\vec{A}, \vec{B}, \vec{C}] = \vec{A} \cdot (\vec{b} \times \vec{c} + \vec{a} \times \vec{b} + 2\vec{a} \times \vec{c}) \]

### Step 6: Expand the dot product
Expanding this gives:
\[ = \vec{A} \cdot (\vec{b} \times \vec{c}) + \vec{A} \cdot (\vec{a} \times \vec{b}) + 2\vec{A} \cdot (\vec{a} \times \vec{c}) \]

### Step 7: Evaluate each term
Substituting \( \vec{A} \):
\[ = (\vec{a} - 2\vec{b} + \vec{c}) \cdot (\vec{b} \times \vec{c}) + (\vec{a} - 2\vec{b} + \vec{c}) \cdot (\vec{a} \times \vec{b}) + 2(\vec{a} - 2\vec{b} + \vec{c}) \cdot (\vec{a} \times \vec{c}) \]

Using the property that the dot product of a vector with a cross product of two other vectors is zero if the vector is one of the two being crossed:
- \( \vec{a} \cdot (\vec{b} \times \vec{c}) \) remains.
- \( -2\vec{b} \cdot (\vec{b} \times \vec{c}) = 0 \).
- \( \vec{c} \cdot (\vec{b} \times \vec{c}) = 0 \).

Thus, we find:
\[ = \vec{a} \cdot (\vec{b} \times \vec{c}) - 2\vec{b} \cdot (\vec{a} \times \vec{b}) + 2\vec{c} \cdot (\vec{a} \times \vec{c}) \]

### Step 8: Final simplification
All terms involving \( \vec{b} \) and \( \vec{c} \) with their respective cross products will yield zero, leading to:
\[ = 0 \]

### Conclusion
Thus, the final answer is:
\[ [\vec{A}, \vec{B}, \vec{C}] = 0 \]