$A$ unit vector in the plane of the vectors $2i + j + k$ and $i - j + k$ and orthogonal to $5i + 2j + 6k$ is

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} - \hat{k}$,$\overrightarrow{b} = \hat{i} - \hat{j}$ and $\overrightarrow{c} = \hat{i} - \hat{j} - \hat{k}$ be three given vectors. If $\overrightarrow{r}$ is a vector such that $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{c} \times \overrightarrow{a}$ and $\overrightarrow{r} \cdot \overrightarrow{b} = 0$,then $\overrightarrow{r} \cdot \overrightarrow{a}$ is equal to ...........

If $a \cdot i = 4$,then $(a \times j) \cdot (2j - 3k) = $

$\vec{a}, \vec{b}, \vec{c}$ are three unit vectors such that $|\vec{a}+\vec{b}+\vec{c}|=1$ and $\vec{a}$ is perpendicular to $\vec{b}$. If $\vec{c}$ makes angles $\alpha, \beta$ with $\vec{a}, \vec{b}$ respectively, then $\cos \alpha+\cos \beta=$

If the position vectors of the vertices $A, B$,and $C$ of a triangle $ABC$ are $4\hat{i} + 7\hat{j} + 8\hat{k}$,$2\hat{i} + 3\hat{j} + 4\hat{k}$,and $2\hat{i} + 5\hat{j} + 7\hat{k}$ respectively,then the position vector of the point where the bisector of angle $A$ meets $BC$ is:

Given the following simultaneous equations for vectors $x$ and $y$:
$(i) x + y = a$
$(ii) x \times y = b$
$(iii) x \cdot a = 1$
Then $x = ?, y = ?$