If $(2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + \lambda \hat{j} + \mu \hat{k}) = 0$,then $\lambda + \mu =$ . . . . . . .

The magnitude of the projection of the vector $2\hat{i} + 3\hat{j} + \hat{k}$ on the vector perpendicular to the plane containing the vectors $\hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + 2\hat{j} + 3\hat{k}$ is

If $\overrightarrow{A}=i-2j-3k,\,\overrightarrow{B}=2i+j-k,\,\overrightarrow{C}=i+3j-2k$,then $(\overrightarrow A \times \overrightarrow B ) \times \overrightarrow C $ is

The area of the parallelogram whose diagonals are the vectors $2\vec{a} - \vec{b}$ and $4\vec{a} - 5\vec{b},$ where $\vec{a}$ and $\vec{b}$ are unit vectors forming an angle of $45^{\circ},$ is

If $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$,find $\vec{a} \times \vec{b}$.

Let $\vec{a} = 3\hat{i} + 2\hat{j} + x\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$,for some real $x$. Then $|\vec{a} \times \vec{b}| = r$ is possible if