The area of the triangle,whose vertices are $A \equiv(1,-1,2)$,$B \equiv(2,1,-1)$ and $C \equiv(3,-1,2)$,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:

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

Find a unit vector perpendicular to the vector $2\hat{i} - \hat{j} + 2\hat{k}$ and coplanar with the vectors $\hat{i} + 2\hat{j} - \hat{k}$ and $2\hat{i} + \hat{j} - \hat{k}$.

If the magnitude of the vector product of the vector $\hat{i}+\hat{j}+\hat{k}$ with a unit vector along the sum of the vectors $2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\lambda \hat{i}+2 \hat{j}+3 \hat{k}$ is equal to $\sqrt{2}$,then the value of ' $\lambda$ ' is

Let $\hat{a}$ be a unit vector perpendicular to the vectors $\overrightarrow{b} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\overrightarrow{c} = 2\hat{i} + 3\hat{j} - \hat{k}$,and makes an angle of $\cos^{-1}\left(-\frac{1}{3}\right)$ with the vector $\hat{i} + \hat{j} + \hat{k}$. If $\hat{a}$ makes an angle of $\frac{\pi}{3}$ with the vector $\hat{i} + \alpha\hat{j} + \hat{k}$,then the value of $\alpha$ is: