For any three non-zero vectors vec{r}_{1}, ve | vedclass

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(D) The given determinant is the Gram determinant $G(\vec{r}_{1}, \vec{r}_{2}, \vec{r}_{3})$,which is equal to the square of the scalar triple product

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All the three vectors are perpendicular to each other.

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(D) The given determinant is the Gram determinant $G(\vec{r}_{1}, \vec{r}_{2}, \vec{r}_{3})$,which is equal to the square of the scalar triple product: $G(\vec{r}_{1}, \vec{r}_{2}, \vec{r}_{3}) = [\vec{r}_{1} \vec{r}_{2} \vec{r}_{3}]^2$. Given that the determinant is $0$,it implies that $[\vec{r}_{1} \vec{r}_{2} \vec{r}_{3}] = 0$. This condition means that the three vectors are coplanar,which implies they are linearly dependent. If vectors are linearly dependent,they lie in the same plane. Option $D$ states that all three vectors are perpendicular to each other,which would imply they are linearly independent (forming an orthogonal basis in $3D$ space),so the determinant would be non-zero. Therefore,the statement in option $D$ is false.

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