If $A(1,-1,2)$,$B(5,7,-6)$,$C(3,4,-10)$,and $D(-1,-4,-2)$ are the vertices of a quadrilateral $ABCD$,then its area is:

$A$ vector of length $3$ perpendicular to each of the vectors $3\,i + j - 4\,k$ and $6\,i + 5\,j - 2\,k$ is

Let $\bar{a} = \hat{i} + \hat{j} - \hat{k}$ and $\bar{c} = 5\hat{i} - 3\hat{j} + 2\hat{k}$. If $\bar{b} \times \bar{c} = \bar{a}$,then find $|\bar{b}|$.

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

The vector of magnitude $6$ units and perpendicular to vectors $2 \hat{i}+\hat{j}-3 \hat{k}$ and $\hat{i}-2 \hat{j}+\hat{k}$ is

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