The area of the parallelogram whose diagonals are $a = 3i + j - 2k$ and $b = i - 3j + 4k$ is

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{c}=\hat{j}-\hat{k}$,$\vec{a} \times \vec{b}=\vec{c}$ and $\vec{a} \cdot \vec{b}=1$,then $\vec{b}$ is equal to:

If $|\vec{a}|=3$,then the value of $|\vec{a} \times \hat{i}|^2+|\vec{a} \times \hat{j}|^2+|\vec{a} \times \hat{k}|^2$ is . . . . . . .

Let $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ be the position vectors of the vertices $A, B, C$ respectively of $\triangle ABC$. The vector area of $\triangle ABC$ is:

The area (in sq. units) of the parallelogram whose diagonals are along the vectors $8\hat{i} - 6\hat{j}$ and $3\hat{i} + 4\hat{j} - 12\hat{k}$ is:

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