Let `veca vecb and vecc` be non- zero vectors aned `vecV_(1) =veca xx (vecb xx vecc) and vecV_(2) = (veca xx vecb) xx vecc`.vectors `vecV_(1) and vecV_(2)` are equal . Then

Options

A. `veca and vecb` ar orthogonal
B. `veca and vecc` are collinear
C. `vecb and vecc` ar orthogonal
D. `vecb= lambda (veca xx vecc) " when " lambda ` is a scalar

Explanation

To solve the problem, we need to analyze the given vectors \( \vec{V_1} \) and \( \vec{V_2} \) and use the properties of vector products.

### Step-by-Step Solution:

1. **Define the Vectors**:
We have two vectors defined as follows:
\[ \vec{V_1} = \vec{a} \times (\vec{b} \times \vec{c}) \]
\[ \vec{V_2} = (\vec{a} \times \vec{b}) \times \vec{c} \]

2. **Set the Vectors Equal**:
Since it is given that \( \vec{V_1} = \vec{V_2} \), we can write:
\[ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \times \vec{c} \]

3. **Use the Vector Triple Product Identity**:
We can apply the vector triple product identity, which states:
\[ \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z} \]
Applying this to \( \vec{V_1} \):
\[ \vec{V_1} = \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \]

4. **Apply the Identity to \( \vec{V_2} \)**:
Now, applying the identity to \( \vec{V_2} \):
\[ \vec{V_2} = (\vec{a} \times \vec{b}) \times \vec{c} = (\vec{c} \cdot \vec{a}) \vec{b} - (\vec{c} \cdot \vec{b}) \vec{a} \]

5. **Set the Expanded Forms Equal**:
Now we have:
\[ (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = (\vec{c} \cdot \vec{a}) \vec{b} - (\vec{c} \cdot \vec{b}) \vec{a} \]

6. **Rearranging the Equation**:
Rearranging gives us:
\[ (\vec{a} \cdot \vec{c}) \vec{b} + (\vec{c} \cdot \vec{b}) \vec{a} = (\vec{a} \cdot \vec{b}) \vec{c} + (\vec{c} \cdot \vec{a}) \vec{b} \]

7. **Analyzing the Coefficients**:
From this equation, we can derive two conditions:
- If \( \vec{a} \cdot \vec{b} = 0 \), then \( \vec{a} \) and \( \vec{b} \) are orthogonal.
- If \( \vec{c} \) is a scalar multiple of \( \vec{a} \) (collinear), then \( \vec{c} = k\vec{a} \) for some scalar \( k \).

8. **Conclusion**:
The conditions derived indicate that:
- \( \vec{a} \) and \( \vec{c} \) are collinear.
- \( \vec{b} \) can be expressed as a linear combination of \( \vec{a} \) and \( \vec{c} \).

### Final Result:
Thus, the correct options are:
- \( \vec{a} \) and \( \vec{c} \) are collinear.
- \( \vec{b} \) can be expressed as \( \vec{b} = \lambda (\vec{a} \times \vec{c}) \) for some scalar \( \lambda \).