If $A(3,2,-1), B(-2,2,-3)$ and $D(-2,5,-4)$ are the vertices of a parallelogram,then the area of the parallelogram is

For any three vectors $\vec{a}, \vec{b}, \vec{c}$,the value of $\vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times (\vec{c} + \vec{a}) + \vec{c} \times (\vec{a} + \vec{b})$ 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:

The area of the triangle whose vertices are $A(1, 1, 1)$,$B(1, 2, 3)$,and $C(2, 3, 1)$ is . . . . . . .

Let $\bar{a}=\bar{i}-\bar{j}+\bar{k}, \bar{b}=\bar{i}-2\bar{j}-2\bar{k}, \bar{c}=6\bar{i}+3\bar{j}-2\bar{k}$ be three vectors. If $\bar{d}$ is a vector perpendicular to both $\bar{a}$ and $\bar{b}$, and $|\bar{d} \times \bar{c}|=14$, then $|\bar{d} \cdot \bar{c}|=$

$A$ vector which is orthogonal to the vector $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and coplanar with the vectors $\vec{b} = 3\hat{i} + 2\hat{j}$ and $\vec{c} = 2\hat{i} + \hat{j} + 3\hat{k}$ is