If $\hat{i}$ is the position vector of the centroid $G$ of triangle $ABC$ and $2\hat{i}+\hat{j}+\hat{k}$ and $2\hat{i}+4\hat{j}-4\hat{k}$ are respectively the position vectors of its vertices $A$ and $B$,then $AG^2+BG^2+CG^2=$

Let $u, v$ and $w$ be non-coplanar vectors. Then the points corresponding to which of the following vectors are collinear?

In the given figure (a square),identify the co-initial vectors.

Compute the magnitude of the following vectors:
$\vec{a}=\hat{i}+\hat{j}+\hat{k} ; \quad \vec{b}=2 \hat{i}-7 \hat{j}-3 \hat{k} ; \quad \vec{c}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}$

Let $\vec{\alpha} = (\lambda - 2) \vec{a} + \vec{b}$ and $\vec{\beta} = (4\lambda - 2)\vec{a} + 3\vec{b}$ be two given vectors where $\vec{a}$ and $\vec{b}$ are non-collinear. The value of $\lambda$ for which vectors $\vec{\alpha}$ and $\vec{\beta}$ are collinear is:

The magnitude of the sum of vectors $(1, -\sqrt{2})$ and $(2, \sqrt{2})$ is $\ldots$