Let `vecr, veca, vecb` and `vecc` be four non-zero vectors such that `vecr.veca=0, |vecrxxvecb|=|vecr||vecb|,|vecrxxvecc|=|vecr||vecc|` then
`[(veca, vecb, vecc)]=`

Options

A. |a||b||c|
B. `-|a||b||c|`
C. 0
D. none of these

Explanation

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

### Step-by-Step Solution:

1. **Given Conditions**:
- \(\vec{r} \cdot \vec{a} = 0\)
- \(|\vec{r} \times \vec{b}| = |\vec{r}| |\vec{b}|\)
- \(|\vec{r} \times \vec{c}| = |\vec{r}| |\vec{c}|\)

2. **Interpret the First Condition**:
- The condition \(\vec{r} \cdot \vec{a} = 0\) implies that the vectors \(\vec{r}\) and \(\vec{a}\) are perpendicular. Therefore, we can conclude:
\[ \text{Angle between } \vec{r} \text{ and } \vec{a} = 90^\circ \]

3. **Interpret the Second Condition**:
- The condition \(|\vec{r} \times \vec{b}| = |\vec{r}| |\vec{b}|\) implies that the angle between \(\vec{r}\) and \(\vec{b}\) is \(90^\circ\) (since the sine of \(90^\circ\) is 1). Therefore, we can conclude:
\[ \text{Angle between } \vec{r} \text{ and } \vec{b} = 90^\circ \]

4. **Interpret the Third Condition**:
- Similarly, the condition \(|\vec{r} \times \vec{c}| = |\vec{r}| |\vec{c}|\) implies that the angle between \(\vec{r}\) and \(\vec{c}\) is also \(90^\circ\). Therefore, we can conclude:
\[ \text{Angle between } \vec{r} \text{ and } \vec{c} = 90^\circ \]

5. **Conclusion about Coplanarity**:
- Since \(\vec{r}\) is perpendicular to \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), we can conclude that \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) must lie in the same plane (i.e., they are coplanar).

6. **Calculate the Scalar Triple Product**:
- The scalar triple product \([\vec{a}, \vec{b}, \vec{c}]\) represents the volume of the parallelepiped formed by the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). If the vectors are coplanar, this volume is zero:
\[ [\vec{a}, \vec{b}, \vec{c}] = 0 \]

### Final Answer:
\[ [\vec{a}, \vec{b}, \vec{c}] = 0 \]