Let `vec(a), vec(b), vec(c)` be non-coplanar vectors and `vec(p)=(vec(b)xxvec(c))/([vec(a)vec(b)vec(c)]), vec(q)=(vec(c)xxvec(a))/([vec(a)vec(b)vec(c)]), vec(r)=(vec(a)xxvec(b))/([vec(a)vec(b)vec(c)])`.
What is the value of
`(vec(a)-vec(b)-vec(c)).vec(p)+(vec(b)-vec(c)-vec(a)).vec(q)+(vec(c)-vec(a)-vec(b)).vec(r)` ?

Options

A. `0`
B. `-3`
C. `3`
D. `-9`

Explanation

To solve the problem, we need to evaluate the expression:

\[ (\vec{a} - \vec{b} - \vec{c}) \cdot \vec{p} + (\vec{b} - \vec{c} - \vec{a}) \cdot \vec{q} + (\vec{c} - \vec{a} - \vec{b}) \cdot \vec{r} \]

where

\[ \vec{p} = \frac{\vec{b} \times \vec{c}}{[\vec{a} \vec{b} \vec{c}]}, \quad \vec{q} = \frac{\vec{c} \times \vec{a}}{[\vec{a} \vec{b} \vec{c}]}, \quad \vec{r} = \frac{\vec{a} \times \vec{b}}{[\vec{a} \vec{b} \vec{c}]} \]

### Step 1: Substitute the values of \(\vec{p}\), \(\vec{q}\), and \(\vec{r}\)

Substituting the values of \(\vec{p}\), \(\vec{q}\), and \(\vec{r}\) into the expression gives:

\[ (\vec{a} - \vec{b} - \vec{c}) \cdot \left(\frac{\vec{b} \times \vec{c}}{[\vec{a} \vec{b} \vec{c}]}\right) + (\vec{b} - \vec{c} - \vec{a}) \cdot \left(\frac{\vec{c} \times \vec{a}}{[\vec{a} \vec{b} \vec{c}]}\right) + (\vec{c} - \vec{a} - \vec{b}) \cdot \left(\frac{\vec{a} \times \vec{b}}{[\vec{a} \vec{b} \vec{c}]}\right) \]

### Step 2: Factor out the common denominator

The common denominator is \([\vec{a} \vec{b} \vec{c}]\), so we can factor it out:

\[ \frac{1}{[\vec{a} \vec{b} \vec{c}]} \left[ (\vec{a} - \vec{b} - \vec{c}) \cdot (\vec{b} \times \vec{c}) + (\vec{b} - \vec{c} - \vec{a}) \cdot (\vec{c} \times \vec{a}) + (\vec{c} - \vec{a} - \vec{b}) \cdot (\vec{a} \times \vec{b}) \right] \]

### Step 3: Evaluate each dot product

1. **First term**:
\[ (\vec{a} - \vec{b} - \vec{c}) \cdot (\vec{b} \times \vec{c}) = \vec{a} \cdot (\vec{b} \times \vec{c}) - \vec{b} \cdot (\vec{b} \times \vec{c}) - \vec{c} \cdot (\vec{b} \times \vec{c}) \]
The second and third terms are zero because the dot product of a vector with a cross product of itself and another vector is zero. Thus, this simplifies to:
\[ \vec{a} \cdot (\vec{b} \times \vec{c}) = [\vec{a} \vec{b} \vec{c}] \]

2. **Second term**:
\[ (\vec{b} - \vec{c} - \vec{a}) \cdot (\vec{c} \times \vec{a}) = \vec{b} \cdot (\vec{c} \times \vec{a}) - \vec{c} \cdot (\vec{c} \times \vec{a}) - \vec{a} \cdot (\vec{c} \times \vec{a}) \]
Again, the second and third terms are zero, so this simplifies to:
\[ \vec{b} \cdot (\vec{c} \times \vec{a}) = [\vec{a} \vec{b} \vec{c}] \]

3. **Third term**:
\[ (\vec{c} - \vec{a} - \vec{b}) \cdot (\vec{a} \times \vec{b}) = \vec{c} \cdot (\vec{a} \times \vec{b}) - \vec{a} \cdot (\vec{a} \times \vec{b}) - \vec{b} \cdot (\vec{a} \times \vec{b}) \]
The second and third terms are zero, so this simplifies to:
\[ \vec{c} \cdot (\vec{a} \times \vec{b}) = [\vec{a} \vec{b} \vec{c}] \]

### Step 4: Combine the results

Now we can combine all the results:

\[ [\vec{a} \vec{b} \vec{c}] + [\vec{a} \vec{b} \vec{c}] + [\vec{a} \vec{b} \vec{c}] = 3[\vec{a} \vec{b} \vec{c}] \]

### Step 5: Final expression

Putting it all together, we have:

\[ \frac{3[\vec{a} \vec{b} \vec{c}]}{[\vec{a} \vec{b} \vec{c}]} = 3 \]

Thus, the final value is:

\[ \boxed{3} \]