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 $a(\vec{\alpha} \times \vec{\beta}) + b(\vec{\beta} \times \vec{\gamma}) + c(\vec{\gamma} \times \vec{\alpha}) = \overrightarrow{0}$, where $a, b, c$ are non-zero scalars, then the vectors $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are

Let $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ with $a_{i} > 0$ for $i = 1, 2, 3$ be a vector that makes equal angles with the coordinate axes $OX$,$OY$,and $OZ$. Also,let the projection of $\vec{a}$ on the vector $3 \hat{i} + 4 \hat{j}$ be $7$. Let $\vec{b}$ be a vector obtained by rotating $\vec{a}$ by $90^{\circ}$. If $\vec{a}$,$\vec{b}$,and the $x$-axis are coplanar,then the projection of vector $\vec{b}$ on $3 \hat{i} + 4 \hat{j}$ is equal to

Let $\bar{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$,where $a_1, a_2, a_3$ and $|\bar{a}|$ are rational numbers. If $\bar{a}$ makes an angle of $45^{\circ}$ with $\bar{b} = \sqrt{2} \hat{i} + 3 \sqrt{2} \hat{j} + 4 \hat{k}$,then $\bar{a}$ lies in:

The vector $2\hat{i} + a\hat{j} + \hat{k}$ is perpendicular to the vector $2\hat{i} - \hat{j} - \hat{k},$ if $a = $

If $\bar{a}, \bar{b}, \bar{c}$ are perpendicular to $\bar{b}+\bar{c}, \bar{c}+\bar{a}$ and $\bar{a}+\bar{b}$ respectively and $|\bar{a}+\bar{b}|=2, |\bar{b}+\bar{c}|=6, |\bar{c}+\bar{a}|=4$,then $|\bar{a}+\bar{b}+\bar{c}|=$