Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of the vertices of a triangle $ABC$. Through the vertices,lines are drawn parallel to the sides to form the triangle $A'B'C'$. Then the centroid of $\Delta A'B'C'$ is

Let $\vec{a} = \sqrt{7}\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{k}$. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}$ and $\vec{r} \cdot \vec{a} = 0$, then $|3\vec{r}|^2$ is equal to:

Coordinate planes and the planes $\pi_1, \pi_2, \pi_3$ which are respectively parallel to $YZ, ZX, XY$ planes at distances $a, b, c$,form a rectangular parallelepiped. $d_1$ is a diagonal of the face on the $XY$-plane not passing through the origin,and $d_2$ is a diagonal of plane $\pi_2$ coterminous with $d_1$. If none of the coordinates of the vertices of the parallelepiped are negative and the angle between $d_1$ and $d_2$ is $\theta$,then $\cos \theta=$

For a unit vector $\vec{a}$,if $(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a}) = 15$,then find $|\vec{x}|$.

If $a$ and $b$ are vectors such that $|a+b| = |a-b|$,then the angle between $a$ and $b$ is (in $^{\circ}$)

Let $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ be the position vector of a point $A$. Let $\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$ and $\vec{c}=\hat{i}+\hat{j}-2 \hat{k}$ be two vectors and $\vec{r}$ be a vector passing through the point $A$ with position vector $\vec{a}$ and parallel to the vector $\vec{b}$. If the projection of $\vec{r}$ on $\vec{c}$ is $\frac{9}{\sqrt{6}}$, then find $|\vec{r}|$.