If the vectors $\hat{i}-3 \hat{j}+2 \hat{k}$ and $-\hat{i}+2 \hat{j}$ represent the diagonals of a parallelogram,then its area will be

For $\vec{a}$ and $\vec{b}$,$|\vec{a}|=3$,$|\vec{b}|=\frac{\sqrt{2}}{3}$ and $\vec{a} \times \vec{b}$ is a unit vector,then the angle between $\vec{a}$ and $\vec{b}$ is . . . . . . .

The line passing through the point $i + 3j + 2k$ and perpendicular to the lines $r = (i + 2j - k) + \lambda (2i + j + k)$ and $r = (2i + 6j + k) + \mu (i + 2j + 3k)$ 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 $a = i + j + 2k$ and $b = 3i + j + k$,then the value of $a \times b$ is:

If the direction ratios of two lines $L_1$ and $L_2$ are given by $(1, -2, 2)$ and $(-2, 3, -6)$ respectively,then the direction ratios of the line which is perpendicular to the lines $L_1$ and $L_2$ are