If the vertices of a triangle are $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$,then the area of the triangle is:

The unit vectors orthogonal to $3 \hat{i}+2 \hat{j}+6 \hat{k}$ and coplanar with $2 \hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ are

The adjacent sides of a parallelogram are $\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$. Then,the area of the parallelogram is . . . . . . sq. units.

If $a = 2i - 3j - k$ and $b = i + 4j - 2k$,then $a \times b$ is

$A$ unit vector which is coplanar to vectors $i + j + 2k$ and $i + 2j + k$ and perpendicular to $i + j + k$ is

If the position vectors of three points $A, B$ and $C$ are respectively $i + j + k, 2i + 3j - 4k$ and $7i + 4j + 9k$,then the unit vector perpendicular to the plane containing the triangle $ABC$ is