If `veca, vecb, vecc` are any three non coplanar vectors, then `[(veca+vecb+vecc, veca-vecc, veca-vecb)]` is equal to

Options

A. `0`
B. `[(veca, vecb, vecc)]`
C. `2[(veca, vecb, vecc)]`
D. `=3[(veca, vecb, vecc)]`

Explanation

To solve the problem, we need to evaluate the scalar triple product \([( \vec{a} + \vec{b} + \vec{c}, \vec{a} - \vec{c}, \vec{a} - \vec{b})]\). We will follow the steps outlined below:

### Step 1: Rewrite the Scalar Triple Product
We start with the expression:
\[ [( \vec{a} + \vec{b} + \vec{c}, \vec{a} - \vec{c}, \vec{a} - \vec{b})] \]
This can be rewritten using the properties of the scalar triple product:
\[ = (\vec{a} + \vec{b} + \vec{c}) \cdot ((\vec{a} - \vec{c}) \times (\vec{a} - \vec{b})) \]

### Step 2: Expand the Cross Product
Next, we need to compute the cross product \((\vec{a} - \vec{c}) \times (\vec{a} - \vec{b})\):
\[ = \vec{a} \times \vec{a} - \vec{a} \times \vec{b} - \vec{c} \times \vec{a} + \vec{c} \times \vec{b} \]
Since \(\vec{a} \times \vec{a} = \vec{0}\), we simplify this to:
\[ = - \vec{a} \times \vec{b} - \vec{c} \times \vec{a} + \vec{c} \times \vec{b} \]

### Step 3: Substitute Back into the Scalar Triple Product
Now we substitute this back into our expression:
\[ = (\vec{a} + \vec{b} + \vec{c}) \cdot (- \vec{a} \times \vec{b} - \vec{c} \times \vec{a} + \vec{c} \times \vec{b}) \]

### Step 4: Distribute the Dot Product
Distributing the dot product gives us:
\[ = -(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} \times \vec{b}) - (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{c} \times \vec{a}) + (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{c} \times \vec{b}) \]

### Step 5: Evaluate Each Term
Using the properties of the scalar triple product:
1. \((\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} \times \vec{b}) = 0\) (since \(\vec{a} \times \vec{b}\) is perpendicular to \(\vec{a}\) and \(\vec{b}\))
2. \((\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{c} \times \vec{a}) = 0\) (since \(\vec{c} \times \vec{a}\) is perpendicular to \(\vec{c}\) and \(\vec{a}\))
3. \((\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{c} \times \vec{b}) = 0\) (since \(\vec{c} \times \vec{b}\) is perpendicular to \(\vec{c}\) and \(\vec{b}\))

### Step 6: Combine the Results
Since all terms evaluate to zero, we have:
\[ = 0 \]

### Final Result
Thus, the value of the scalar triple product is:
\[ \boxed{0} \]