If $\vec{a}=2 \hat{i}-\hat{j}+3 \hat{k}, \vec{b}=-3 \hat{i}+5 \hat{j}-4 \hat{k}$ and $\vec{c}=6 \hat{i}-4 \hat{j}+5 \hat{k}$,then $(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})=$

The area of a parallelogram whose adjacent sides are $i - 2j + 3k$ and $2i + j - 4k$ is:

$A$ unit vector perpendicular to the plane containing the vectors $\hat{i}+2\hat{j}+\hat{k}$ and $-2\hat{i}+\hat{j}+3\hat{k}$ 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

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{c}=\hat{j}-\hat{k}$ are given vectors,then a vector $\vec{b}$ satisfying the equations $\vec{a} \times \vec{b}=\vec{c}$ and $\vec{a} \cdot \vec{b}=3$ 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: