Let $\bar{a}$,$\bar{b}$,and $\bar{c}$ be unit vectors. Suppose that $\bar{a} \cdot \bar{b} = \bar{a} \cdot \bar{c} = 0$ and the angle between $\bar{b}$ and $\bar{c}$ is $\frac{\pi}{6}$. Then $\bar{a}$ is equal to:

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:

One side and one diagonal of a parallelogram are represented by $3 \hat{i}+\hat{j}-\hat{k}$ and $2 \hat{i}+\hat{j}-2 \hat{k}$ respectively. Then,the area of the parallelogram in square units is:

If $\overrightarrow{a} = \hat{i} + \hat{j} + \hat{k}$,$\overrightarrow{b} = \hat{i} + 3\hat{j} + 5\hat{k}$ and $\overrightarrow{c} = 7\hat{i} + 9\hat{j} + 11\hat{k}$,then the area of the parallelogram having diagonals $\overrightarrow{a} + \overrightarrow{b}$ and $\overrightarrow{b} + \overrightarrow{c}$ is:

Let $L_1: \overrightarrow{r}=(\hat{i}-\hat{j}+2 \hat{k})+\lambda(\hat{i}-\hat{j}+2 \hat{k}), \lambda \in R$,$L_2: \overrightarrow{r}=(\hat{j}-\hat{k})+\mu(3 \hat{i}+\hat{j}+p \hat{k}), \mu \in R$,and $L_3: \overrightarrow{r}=\delta(\ell \hat{i}+m \hat{j}+n \hat{k}), \delta \in R$ be three lines such that $L_1$ is perpendicular to $L_2$ and $L_3$ is perpendicular to both $L_1$ and $L_2$. Then the point which lies on $L_3$ is

Let $A(2, 3, 5)$ and $C(-3, 4, -2)$ be opposite vertices of a parallelogram $ABCD$. If the diagonal $\overrightarrow{BD} = \hat{i} + 2\hat{j} + 3\hat{k}$,then the area of the parallelogram is equal to: