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

Let $\overrightarrow{a}=\hat{i}+5\hat{j}+\alpha\hat{k}$,$\overrightarrow{b}=\hat{i}+3\hat{j}+\beta\hat{k}$ and $\overrightarrow{c}=-\hat{i}+2\hat{j}-3\hat{k}$ be three vectors such that $|\overrightarrow{b} \times \overrightarrow{c}|=5\sqrt{3}$ and $\overrightarrow{a}$ is perpendicular to $\overrightarrow{b}$. Then the greatest value of $|\vec{a}|^{2}$ is .... .

Let $\hat{u} = u_1 \hat{i} + u_2 \hat{j} + u_3 \hat{k}$ be a unit vector in $\mathbb{R}^3$ and $\hat{v} = \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} + 2 \hat{k})$. Given that there exists a unit vector $\vec{w}$ such that $\hat{u} \times \vec{w} = \hat{v}$,which of the following is(are) correct?

Let $\vec{a}=\hat{i}+2 \hat{j}+\lambda \hat{k}$,$\vec{b}=3 \hat{i}-5 \hat{j}-\lambda \hat{k}$,$\vec{a} \cdot \vec{c}=7$,$2 \vec{b} \cdot \vec{c}+43=0$,and $\vec{a} \times \vec{c}=\vec{b} \times \vec{c}$. Then $|\vec{a} \cdot \vec{b}|$ is equal to

Consider the lines $L_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $L_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$. Then the unit vector perpendicular to both $L_1$ and $L_2$ is:

If $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}$ represent the adjacent sides of a parallelogram,then the area of this parallelogram is: