The position vectors of the points $A, B, C$ are $(2\hat{i}+\hat{j}-\hat{k}), (3\hat{i}-2\hat{j}+\hat{k})$ and $(\hat{i}+4\hat{j}-3\hat{k})$ respectively. These points

$(r \cdot i)^2 + (r \cdot j)^2 + (r \cdot k)^2 = $

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $\vec{a} \cdot \vec{a} = \vec{b} \cdot \vec{b} = \vec{c} \cdot \vec{c} = 5$ and $|\vec{a} + \vec{b} - \vec{c}|^2 + |\vec{b} + \vec{c} - \vec{a}|^2 + |\vec{c} + \vec{a} - \vec{b}|^2 = 50$. Then $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = $

Let $\vec{a}=x \hat{i}+y \hat{j}+z \hat{k}$ and $x=2 y$. If $|\vec{a}|=5 \sqrt{2}$ and $\vec{a}$ makes an angle of $135^{\circ}$ with the $z$-axis,then $\vec{a}=$

Answer the following as true or false.
$\vec{a}$ and $-\vec{a}$ are collinear.

The vector in the direction of vector $5 \hat{i} - \hat{j} + 2 \hat{k}$ with a magnitude of $8$ units is: