If $a$ is a unit vector, then $|a \times \hat{i}|^2+|a \times \hat{j}|^2+|a \times \hat{k}|^2=$

If the position vectors of the vertices $A, B$ and $C$ of $\triangle ABC$ are $\hat{i}+2\hat{j}-5\hat{k}$,$-2\hat{i}+2\hat{j}+\hat{k}$ and $2\hat{i}+\hat{j}-\hat{k}$ respectively,then $\angle B=$

The value of $b$ such that the scalar product of the vector $(i + j + k)$ with the unit vector parallel to the sum of the vectors $(2i + 4j - 5k)$ and $(bi + 2j + 3k)$ is $1$,is:

Let the arc $AC$ of a circle subtend a right angle at the centre $O$. If the point $B$ on the arc $AC$ divides the arc $AC$ such that $\frac{\text{length of arc } AB}{\text{length of arc } BC} = \frac{1}{5}$,and $\overrightarrow{OC} = \alpha \overrightarrow{OA} + \beta \overrightarrow{OB}$,then $\alpha + \sqrt{2}(\sqrt{3}-1) \beta$ is equal to

Let $\overrightarrow{x}$ be a vector in the plane containing vectors $\overrightarrow{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\overrightarrow{b} = \hat{i} + 2\hat{j} - \hat{k}$. If the vector $\overrightarrow{x}$ is perpendicular to $(3\hat{i} + 2\hat{j} - \hat{k})$ and its projection on $\overrightarrow{a}$ is $\frac{17\sqrt{6}}{2}$,then the value of $|\overrightarrow{x}|^{2}$ is equal to ...... .

Let $a=\hat{i}+2 \hat{j}-2 \hat{k}$ and $b=2 \hat{i}-\hat{j}-2 \hat{k}$. If the orthogonal projection vector of $a$ on $b$ is $x$ and the orthogonal projection vector of $b$ on $a$ is $y$, then $|x-y|=$