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

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:

If $2\vec{a} + 3\vec{b} + \vec{c} = \vec{0}$,then $\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$ is equal to

If $\overrightarrow{a} = \hat{i} + \hat{j} + \hat{k}$,$\overrightarrow{b} = \hat{i} + 3\hat{j} + 5\hat{k}$ and $\overrightarrow{c} = 7\hat{i} + 9\hat{j} + 11\hat{k}$,then the area of the parallelogram having diagonals $\overrightarrow{a} + \overrightarrow{b}$ and $\overrightarrow{b} + \overrightarrow{c}$ is:

If $a$ and $b$ are unit vectors such that $a \times b$ is also a unit vector,then the angle between $a$ and $b$ is

$A$ unit vector in the plane of $i + 2j + k$ and $i + j + 2k$ which is perpendicular to $2i + j + k$ is