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

The unit vector perpendicular to the vectors $6i + 2j + 3k$ and $3i - 6j - 2k$ is

The area of the parallelogram whose adjacent sides are $\vec{a} = \hat{i} - \hat{k}$ and $\vec{b} = 2\hat{j} + 3\hat{k}$ is

Let $\vec{a} = \alpha \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 2 \hat{i} + \hat{j} - \alpha \hat{k}$,where $\alpha > 0$. If the projection of $\vec{a} \times \vec{b}$ on the vector $\vec{c} = -\hat{i} + 2 \hat{j} - 2 \hat{k}$ is $30$,then $\alpha$ is equal to:

If $\vec{u}$ and $\vec{v}$ are unit vectors and $\theta$ is the acute angle between them,then $2\vec{u} \times 3\vec{v}$ is a unit vector for

Vectors $\vec{p}=a \hat{i}+b \hat{j}+c \hat{k}$, $\vec{q}=d \hat{i}+3 \hat{j}+4 \hat{k}$ and $\vec{r}=3 \hat{i}+\hat{j}-2 \hat{k}$ form a triangle $ABC$ such that $\vec{p}=\vec{q}+\vec{r}$. If the area of $\triangle ABC$ is $5 \sqrt{6}$ sq. units, then the sum of the absolute values of $a, b, c$ is