If `vecP = (vecbxxvecc)/([vecavecbvecc]).vecq=(veccxxveca)/([veca vecb vecc])and vecr = (vecaxxvecb)/([veca vecbvecc]), " where " veca,vecb and vecc` are three non- coplanar vectors then the value of the expression `(veca + vecb + vecc ). (vecq+ vecq+vecr)` is

Options

A. 3
B. 2
C. 1
D. 0

Explanation

To solve the problem, we need to evaluate the expression \((\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{q} + \vec{q} + \vec{r})\), where:

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

Here, \([\vec{a}, \vec{b}, \vec{c}]\) denotes the scalar triple product of vectors \(\vec{a}, \vec{b}, \vec{c}\).

### Step 1: Calculate \(\vec{q} + \vec{r}\)

We first need to find \(\vec{q} + \vec{r}\):
\[ \vec{q} + \vec{r} = \frac{\vec{c} \times \vec{a}}{[\vec{a}, \vec{b}, \vec{c}]} + \frac{\vec{a} \times \vec{b}}{[\vec{a}, \vec{b}, \vec{c}]} \]
Combining the fractions:
\[ \vec{q} + \vec{r} = \frac{\vec{c} \times \vec{a} + \vec{a} \times \vec{b}}{[\vec{a}, \vec{b}, \vec{c}]} \]

### Step 2: Substitute into the expression

Now we substitute \(\vec{q} + \vec{r}\) into the expression:
\[ (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{q} + \vec{r}) = (\vec{a} + \vec{b} + \vec{c}) \cdot \left(\frac{\vec{c} \times \vec{a} + \vec{a} \times \vec{b}}{[\vec{a}, \vec{b}, \vec{c}]}\right) \]
This simplifies to:
\[ = \frac{1}{[\vec{a}, \vec{b}, \vec{c}]} \left((\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{c} \times \vec{a}) + (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} \times \vec{b})\right) \]

### Step 3: Calculate each dot product

1. **For \((\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{c} \times \vec{a})\)**:
- This is equal to \(\vec{a} \cdot (\vec{c} \times \vec{a}) + \vec{b} \cdot (\vec{c} \times \vec{a}) + \vec{c} \cdot (\vec{c} \times \vec{a})\).
- The first and last terms are zero because \(\vec{a} \cdot (\vec{c} \times \vec{a}) = 0\) and \(\vec{c} \cdot (\vec{c} \times \vec{a}) = 0\).
- Thus, we have \(\vec{b} \cdot (\vec{c} \times \vec{a})\).

2. **For \((\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} \times \vec{b})\)**:
- This is equal to \(\vec{a} \cdot (\vec{a} \times \vec{b}) + \vec{b} \cdot (\vec{a} \times \vec{b}) + \vec{c} \cdot (\vec{a} \times \vec{b})\).
- The first two terms are zero, so we have \(\vec{c} \cdot (\vec{a} \times \vec{b})\).

### Step 4: Combine results

Now, we combine the results:
\[ = \frac{1}{[\vec{a}, \vec{b}, \vec{c}]} \left(\vec{b} \cdot (\vec{c} \times \vec{a}) + \vec{c} \cdot (\vec{a} \times \vec{b})\right) \]
Using the scalar triple product identity:
\[ \vec{b} \cdot (\vec{c} \times \vec{a}) = [\vec{a}, \vec{b}, \vec{c}] \]
and
\[ \vec{c} \cdot (\vec{a} \times \vec{b}) = [\vec{a}, \vec{b}, \vec{c}] \]
Thus, we have:
\[ = \frac{[\vec{a}, \vec{b}, \vec{c}] + [\vec{a}, \vec{b}, \vec{c}]}{[\vec{a}, \vec{b}, \vec{c}]} = \frac{2[\vec{a}, \vec{b}, \vec{c}]}{[\vec{a}, \vec{b}, \vec{c}]} = 2 \]

### Final Step: Add \(\vec{p}\)

Finally, we need to add \(\vec{p}\):
\[ (\vec{a} + \vec{b} + \vec{c}) \cdot \vec{p} = 1 \]

### Conclusion

Thus, the final value of the expression is:
\[ 1 + 2 = 3 \]

### Answer:
The value of the expression \((\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{q} + \vec{q} + \vec{r})\) is **3**.