If `veca, vecb, vecc` are any three vectors such that `(veca+vecb).vecc=(veca-vecb)vecc=0` then`(vecaxxvecb)xxvecc` is

Options

A. `vec0`
B. `veca`
C. `vecb`
D. none of these

Explanation

To solve the problem, we need to find the value of \((\vec{a} \times \vec{b}) \times \vec{c}\) given the conditions \((\vec{a} + \vec{b}) \cdot \vec{c} = 0\) and \((\vec{a} - \vec{b}) \cdot \vec{c} = 0\).

### Step-by-Step Solution:

1. **Write down the given equations:**
\[ (\vec{a} + \vec{b}) \cdot \vec{c} = 0 \quad \text{(1)} \]
\[ (\vec{a} - \vec{b}) \cdot \vec{c} = 0 \quad \text{(2)} \]

2. **Expand the equations:**
From equation (1):
\[ \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c} = 0 \quad \text{(3)} \]
From equation (2):
\[ \vec{a} \cdot \vec{c} - \vec{b} \cdot \vec{c} = 0 \quad \text{(4)} \]

3. **Add equations (3) and (4):**
\[ (\vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c}) + (\vec{a} \cdot \vec{c} - \vec{b} \cdot \vec{c}) = 0 + 0 \]
This simplifies to:
\[ 2(\vec{a} \cdot \vec{c}) = 0 \]
Therefore, we conclude:
\[ \vec{a} \cdot \vec{c} = 0 \quad \text{(5)} \]

4. **Substitute \(\vec{a} \cdot \vec{c} = 0\) into equation (3):**
From equation (3):
\[ 0 + \vec{b} \cdot \vec{c} = 0 \]
This implies:
\[ \vec{b} \cdot \vec{c} = 0 \quad \text{(6)} \]

5. **Now, we need to find \((\vec{a} \times \vec{b}) \times \vec{c}\):**
We can use the vector triple product identity:
\[ \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z} \]
Here, let \(\vec{x} = \vec{a}\), \(\vec{y} = \vec{b}\), and \(\vec{z} = \vec{c}\):
\[ (\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a} \]

6. **Substituting the values from equations (5) and (6):**
Since \(\vec{a} \cdot \vec{c} = 0\) and \(\vec{b} \cdot \vec{c} = 0\):
\[ (\vec{a} \times \vec{b}) \times \vec{c} = 0 \cdot \vec{b} - 0 \cdot \vec{a} = \vec{0} \]

### Final Result:
\[ (\vec{a} \times \vec{b}) \times \vec{c} = \vec{0} \]