Given three vectors $a=2 \hat{i}-\hat{j}+\hat{k}$,$b=\hat{i}+2 \hat{j}-\hat{k}$,and $c=\hat{i}+\hat{j}-2 \hat{k}$,a vector in the plane of $b$ and $c$ whose projection on $a$ is of magnitude $\sqrt{\frac{2}{3}}$ is

Let $\vec{a}=4 \hat{i}+3 \hat{j}$ and $\vec{b}=3 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\vec{c}$ is a vector such that $\vec{c} \cdot(\vec{a} \times \vec{b})+25=0, \vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})=4$ and the projection of $\vec{c}$ on $\vec{a}$ is $1$. Then,the projection of $\vec{c}$ on $\vec{b}$ equals:

Let the position vectors of the vertices $A, B$ and $C$ of a tetrahedron $ABCD$ be $\hat{i}+2\hat{j}+\hat{k}$,$\hat{i}+3\hat{j}-2\hat{k}$ and $2\hat{i}+\hat{j}-\hat{k}$ respectively. The altitude from the vertex $D$ to the opposite face $ABC$ meets the median line segment through $A$ of the triangle $ABC$ at the point $E$. If the length of $AD$ is $\frac{\sqrt{110}}{3}$ and the volume of the tetrahedron is $\frac{\sqrt{805}}{6\sqrt{2}}$,then the position vector of $E$ 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:

If $P=(0,1,2), Q=(4,-2,1)$ and $O=(0,0,0)$,then $\angle POQ=$

Area of a rectangle having vertices $A, B, C$ and $D$ with position vectors $-\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$, $\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$, $\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}$ and $-\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}$, respectively is . . . . . . .