The magnitude of the projection of vector $\vec{a} = -\hat{i} + 2\hat{j} - \hat{k}$ on the vector $\vec{b} = \hat{i} + 2\hat{j} + 2\hat{k}$ 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

Given $ABC$ is an equilateral triangle of side length $1$ unit and $P$ be any arbitrary point on the circumcircle of triangle $ABC,$ then $|\vec{PA}|^2+|\vec{PB}|^2+|\vec{PC}|^2$ is equal to

If $\theta$ is the angle between the vectors $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$,then:

If $G(\bar{g})$,$H(\bar{h})$,and $P(\bar{p})$ are respectively the centroid,orthocenter,and circumcentre of a triangle and $x \bar{p} + y \bar{h} + z \bar{g} = \overline{0}$,then $x, y, z$ are respectively:

Let $ABC$ be a triangle such that $\overrightarrow{BC} = \overrightarrow{a}$,$\overrightarrow{CA} = \overrightarrow{b}$,$\overrightarrow{AB} = \overrightarrow{c}$,$|\overrightarrow{a}| = 6\sqrt{2}$,$|\overrightarrow{b}| = 2\sqrt{3}$,and $\overrightarrow{b} \cdot \overrightarrow{c} = 12$. Consider the statements:
$(S1): |(\overrightarrow{a} \times \overrightarrow{b}) + (\overrightarrow{c} \times \overrightarrow{b})| - |\overrightarrow{c}| = 6(2\sqrt{2} - 1)$
$(S2): \angle ABC = \cos^{-1}\left(\sqrt{\frac{2}{3}}\right)$.
Which of the following is true?