If the given vectors $(-bc, b^2 + bc, c^2 + bc)$,$(a^2 + ac, -ac, c^2 + ac)$ and $(a^2 + ab, b^2 + ab, -ab)$ are coplanar,where none of $a, b$ and $c$ is zero,then:

If $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{c} = r\hat{i} + \hat{j} + (2r - 1)\hat{k}$ are three vectors such that $\vec{c}$ is parallel to the plane of $\vec{a}$ and $\vec{b}$,then $r$ is equal to

Value of $\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{i} \cdot(\hat{j} \times \hat{j})+\hat{k} \cdot(\hat{j} \times \hat{i})+\hat{i} \cdot(\hat{k} \times \hat{j})$ is . . . . . . .

If three vectors $\vec{a} = 12\hat{i} + 4\hat{j} + 3\hat{k}$,$\vec{b} = 8\hat{i} - 12\hat{j} - 9\hat{k}$,and $\vec{c} = 33\hat{i} - 4\hat{j} - 24\hat{k}$ represent the coterminous edges of a parallelepiped,then its volume is:

$a$ is perpendicular to both $b$ and $c$. The angle between $b$ and $c$ is $\frac{2 \pi}{3}$. If $|a|=2$, $|b|=3$, and $|c|=4$, then $c \cdot (a \times b)$ is equal to (in $\sqrt{3}$)

If $x, y$ and $z$ are non-zero real numbers and $\vec{a}=x \hat{i}+2 \hat{j}, \vec{b}=y \hat{j}+3 \hat{k}$ and $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$ are such that $\vec{a} \times \vec{b}=z \hat{i}-3 \hat{j}+xy \hat{k}$ is not given,but $\vec{a} \times \vec{b}=6 \hat{i}-3 \hat{j}+\hat{k}$ is given as $z \hat{i}-3 \hat{j}+\hat{k}$,then the scalar triple product $[\vec{a} \vec{b} \vec{c}]$ is equal to: