Let `veca, vecb, vecc` be three vectors such that `[(veca, vecb, vecc)]=2`. If `vecr=l(vecbxxvecc)+m(veccxxveca)+n(vecaxxvecb)` be perpendicular to `veca+vecb+vecc` , then the value of `l+m+n` is

Options

A. 2
B. 1
C. 0
D. none of these

Answer

To solve the problem step by step, we need to analyze the given vectors and their relationships.

### Step 1: Understand the Given Information
We have three vectors \(\vec{a}, \vec{b}, \vec{c}\) such that the scalar triple product \([\vec{a}, \vec{b}, \vec{c}] = 2\). This means that the volume of the parallelepiped formed by these vectors is 2.

### Step 2: Define the Vector \(\vec{r}\)
We are given that:
\[ \vec{r} = l(\vec{b} \times \vec{c}) + m(\vec{c} \times \vec{a}) + n(\vec{a} \times \vec{b}) \]
This vector \(\vec{r}\) is perpendicular to the vector sum \(\vec{a} + \vec{b} + \vec{c}\).

### Step 3: Use the Perpendicularity Condition
Since \(\vec{r}\) is perpendicular to \(\vec{a} + \vec{b} + \vec{c}\), we can express this condition mathematically:
\[ \vec{r} \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \]

### Step 4: Expand the Dot Product
Substituting \(\vec{r}\) into the dot product gives:
\[ \left(l(\vec{b} \times \vec{c}) + m(\vec{c} \times \vec{a}) + n(\vec{a} \times \vec{b})\right) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \]
This expands to:
\[ l(\vec{b} \times \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) + m(\vec{c} \times \vec{a}) \cdot (\vec{a} + \vec{b} + \vec{c}) + n(\vec{a} \times \vec{b}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \]

### Step 5: Calculate Each Dot Product
Using the properties of the scalar triple product, we know:
1. \((\vec{b} \times \vec{c}) \cdot \vec{a} = [\vec{a}, \vec{b}, \vec{c}] = 2\)
2. \((\vec{c} \times \vec{a}) \cdot \vec{b} = [\vec{b}, \vec{c}, \vec{a}] = 2\)
3. \((\vec{a} \times \vec{b}) \cdot \vec{c} = [\vec{c}, \vec{a}, \vec{b}] = 2\)

Thus, we can write:
- \((\vec{b} \times \vec{c}) \cdot \vec{b} = 0\) and \((\vec{b} \times \vec{c}) \cdot \vec{c} = 0\)
- Similarly, the other terms will yield zero when dotted with their respective vectors.

### Step 6: Combine the Results
This leads to:
\[ l \cdot 2 + m \cdot 2 + n \cdot 2 = 0 \]
or simplifying:
\[ 2(l + m + n) = 0 \]
Thus, we find:
\[ l + m + n = 0 \]

### Conclusion
The value of \(l + m + n\) is \(0\).