Consider the four points $A(1, -2, -1)$,$B(4, 0, -3)$,$C(1, 2, -1)$,and $D(2, -4, -5)$ in space. If $\vec{b} = \vec{AB}$,$\vec{c} = \vec{AC}$,and $\vec{d} = \vec{AD}$,then find the value of $\frac{[\vec{b} \times \vec{c}, \vec{c} \times \vec{d}, \vec{d} \times \vec{b}]}{[\vec{b}+\vec{c}, \vec{c}+\vec{d}, \vec{d}+\vec{b}]}$.

Let $\overrightarrow{OP} = \frac{\alpha-1}{\alpha} \hat{i} + \hat{j} + \hat{k}$,$\overrightarrow{OQ} = \hat{i} + \frac{\beta-1}{\beta} \hat{j} + \hat{k}$ and $\overrightarrow{OR} = \hat{i} + \hat{j} + \frac{1}{2} \hat{k}$ be three vectors,where $\alpha, \beta \in \mathbb{R} - \{0\}$ and $O$ denotes the origin. If $(\overrightarrow{OP} \times \overrightarrow{OQ}) \cdot \overrightarrow{OR} = 0$ and the point $(\alpha, \beta, 2)$ lies on the plane $3x + 3y - z + l = 0$,then the value of $l$ is:

If the vectors $a$,$b$,and $c$ are coplanar,then $\left|\begin{array}{ccc}a & b & c \\ a \cdot a & a \cdot b & a \cdot c \\ b \cdot a & b \cdot b & b \cdot c\end{array}\right|$ is equal to

If $\bar{a} = \hat{i} - \hat{j}$,$\bar{b} = \hat{j} - \hat{k}$,and $\bar{c} = \hat{k} - \hat{i}$,then a unit vector $\bar{d}$ such that $\bar{a} \cdot \bar{d} = 0$ and $[\bar{b} \bar{c} \bar{d}] = 0$ is:

If the position vectors of the vertices $A, B$ and $C$ are $6i$,$6j$ and $k$ respectively with respect to the origin $O$,then the volume of the tetrahedron $OABC$ is

$(\hat{i} \times \hat{j}) \cdot [(\hat{j} \times \hat{k}) \times (\hat{k} \times \hat{i})]$