Vector $\vec{x}$ is a vector in the direction of $(2, -2, 1)$ and has a magnitude of $6$ units. Vector $\vec{y}$ is a vector in the direction of $(1, 1, -1)$ and has a magnitude of $\sqrt{3}$ units. Then,$|\vec{x} + 2\vec{y}| = $ . . . . . . .

The perimeter of the triangle whose vertices have the position vectors $\hat{i}+\hat{j}+\hat{k}$,$5\hat{i}+3\hat{j}-3\hat{k}$,and $2\hat{i}+5\hat{j}+9\hat{k}$ is

If $PQRST$ is a pentagon,then the resultant of forces $\overline{PQ}, \overline{PT}, \overline{QR}, \overline{SR}, \overline{TS}$ and $\overline{PS}$ is

$A$ vector $\vec{r}$ is inclined at equal angles to the three axes. If the magnitude of $\vec{r}$ is $2 \sqrt{3}$ units,find $\vec{r}$.

If $a = \hat{i} + 4 \hat{j}$, $b = 2 \hat{i} - 2 \hat{j}$, and $c = 5 \hat{i} + 9 \hat{j}$, then $c$ is equal to:

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} = $