If the area of a parallelogram,whose diagonals are $\vec{d_1} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{d_2} = 2\hat{i} + 3\hat{j} + \alpha\hat{k}$,is $\frac{\sqrt{93}}{2}$ sq. unit,then $\alpha = $

If $(\bar{i}+\bar{j}+\bar{k})$,$(\bar{i}+2\bar{j}+3\bar{k})$ and $(2\bar{i}-\bar{j}+\bar{k})$ are the position vectors of the vertices $A$,$B$ and $C$ of $\triangle ABC$ respectively,then the vector equation of the altitude through $A$ is

The points $A(a), B(b), C(c)$ will be collinear if

If $\vec{u} = \vec{a} - \vec{b}$ and $\vec{v} = \vec{a} + \vec{b}$ and $|\vec{a}| = |\vec{b}| = 2$,then $|\vec{u} \times \vec{v}| = ......$

If $A, B, C, D$ are any four points in space, then $|\overrightarrow{AB} \times \overrightarrow{CD} + \overrightarrow{BC} \times \overrightarrow{AD} + \overrightarrow{CA} \times \overrightarrow{BD}|$ is equal to (where $\Delta$ denotes the area of $\Delta ABC$) (in $\Delta$)

Consider the cube in the first octant with sides $OP, OQ$ and $OR$ of length $1$,along the $x$-axis,$y$-axis and $z$-axis,respectively,where $O(0,0,0)$ is the origin. Let $S\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$ be the centre of the cube and $T$ be the vertex of the cube opposite to the origin $O$ such that $S$ lies on the diagonal $OT$. If $\overrightarrow{p} = \overrightarrow{SP}, \overrightarrow{q} = \overrightarrow{SQ}, \overrightarrow{r} = \overrightarrow{SR}$ and $\overrightarrow{t} = \overrightarrow{ST}$,then the value of $|(\overrightarrow{p} \times \overrightarrow{q}) \times (\overrightarrow{r} \times \overrightarrow{t})|$ is: