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

If the area of the triangle with vertices $(1, 2, 0)$,$(1, 0, 2)$ and $(0, x, 1)$ is $\sqrt{6}$ square units,then the value of $x$ 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 position vectors of the points $A, B$ and $C$ are $i + j, j + k$ and $k + i$ respectively. The vector area of the $\Delta ABC = \pm \frac{1}{2} \vec{\alpha}$ where $\vec{\alpha} = $

If $a$ and $b$ are unit vectors,then the vector $(a+b) \times (a \times b)$ is parallel to the vector

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