$A$ unit vector perpendicular to the vectors $4i - j + 3k$ and $-2i + j - 2k$ is

$A$ non-zero vector $\vec{a}$ is parallel to the line of intersection of the planes defined by the vectors $\vec{i}, \vec{i} + \vec{j}$ and $\vec{i} - \vec{j}, \vec{i} + \vec{k}$. The angle between $\vec{a}$ and the vector $\vec{i} - 2\vec{j} + 2\vec{k}$ is .....

Let $\overline{OA} = \vec{a}$,$\overline{OB} = 10\vec{a} + 2\vec{b}$,and $\overline{OC} = \vec{b}$,where $O, A, C$ are non-collinear. Let $p$ be the area of the quadrilateral $OABC$ and $q$ be the area of the parallelogram with adjacent sides $OA$ and $OC$. Then $p/q = \dots$

Let $\overline{a}=\alpha \hat{i}+3 \hat{j}-\hat{k}$,$\overline{b}=3 \hat{i}-\beta \hat{j}+4 \hat{k}$ and $\overline{c}=\hat{i}+2 \hat{j}-2 \hat{k}$,where $\alpha, \beta \in R$,be three vectors. If the projection of $\overline{a}$ on $\overline{c}$ is $\frac{10}{3}$ and $\overline{b} \times \overline{c}=-6 \hat{i}+10 \hat{j}+7 \hat{k}$,then the value of $\alpha^2+\beta^2-\alpha \beta$ is equal to

If the position vectors of the vertices of a $\triangle ABC$ are $\vec{OA} = 3\hat{i} + \hat{j} + 2\hat{k}$,$\vec{OB} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{OC} = 2\hat{i} + 3\hat{j} + \hat{k}$,then the length of the altitude of $\triangle ABC$ drawn from $A$ is

Let $\bar{a}, \bar{b}, \bar{c}$ be vectors such that $\bar{a} \neq \bar{o}, \bar{b} \neq \bar{o}, \bar{a} \times \bar{c} = \bar{b}$ and $\bar{b} \times \bar{c} = \bar{a}$. Then: