If a, b,c are non-coplanar vectors such that `r.a = r.b = r.c =0`, then

Options

A. `r = 0`
B. `[abc] =0`
C. `r ne 0, [abc] =0`
D. `r = 0[abc] ne 0`

Answer

To solve the problem, we need to analyze the given conditions involving the vectors \( \vec{a}, \vec{b}, \vec{c} \) and the vector \( \vec{r} \).

### Step-by-Step Solution:

1. **Understanding the Given Conditions**:
We have three non-coplanar vectors \( \vec{a}, \vec{b}, \vec{c} \) such that:
\[ \vec{r} \cdot \vec{a} = \vec{r} \cdot \vec{b} = \vec{r} \cdot \vec{c} = 0 \]
This means that the vector \( \vec{r} \) is orthogonal (perpendicular) to each of the vectors \( \vec{a}, \vec{b}, \vec{c} \).

**Hint**: Remember that if a vector is orthogonal to another vector, their dot product is zero.

2. **Implications of Non-Coplanarity**:
Since \( \vec{a}, \vec{b}, \vec{c} \) are non-coplanar, it implies that they do not lie in the same plane. The scalar triple product of these vectors, defined as \( \vec{a} \cdot (\vec{b} \times \vec{c}) \), must be non-zero.

**Hint**: The scalar triple product being non-zero indicates that the vectors span a three-dimensional space.

3. **Analyzing the Vector \( \vec{r} \)**:
The condition \( \vec{r} \cdot \vec{a} = 0 \) implies that \( \vec{r} \) lies in the plane formed by the vectors \( \vec{b} \) and \( \vec{c} \). Similarly, \( \vec{r} \) is also in the plane formed by \( \vec{a} \) and \( \vec{c} \), and \( \vec{a} \) and \( \vec{b} \).

**Hint**: A vector that is orthogonal to a set of vectors lies in the space defined by the remaining vectors.

4. **Conclusion**:
Since \( \vec{a}, \vec{b}, \vec{c} \) are non-coplanar and \( \vec{r} \) is orthogonal to all three, the only vector that satisfies this condition is the zero vector \( \vec{r} = \vec{0} \). Therefore, the conclusion is that:
\[ \vec{r} = \vec{0} \]

**Hint**: The zero vector is the only vector that can be orthogonal to any set of vectors.

### Final Answer:
The vector \( \vec{r} \) must be the zero vector, i.e., \( \vec{r} = \vec{0} \).