If $\theta$ is the angle between the vectors $2 \hat{i}-\hat{j}+2 \hat{k}$ and $a \hat{i}+4 \hat{j}+b \hat{k}$ and $\cos \theta=\frac{2}{3}$, then $2(a+b+3)=$

Let $\vec{a}=4 \hat{i}-\hat{j}+\hat{k}$,$\vec{b}=11 \hat{i}-\hat{j}+\hat{k}$,and $\vec{c}$ be a vector such that $(\vec{a}+\vec{b}) \times \vec{c} = \vec{c} \times (-2 \vec{a}+3 \vec{b})$. If $(2 \vec{a}+3 \vec{b}) \cdot \vec{c} = 1670$,then $|\vec{c}|^2$ is equal to:

Given $ABC$ is an equilateral triangle of side length $1$ unit and $P$ be any arbitrary point on the circumcircle of triangle $ABC,$ then $|\vec{PA}|^2+|\vec{PB}|^2+|\vec{PC}|^2$ is equal to

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

If $a=2 \hat{i}+\hat{j}-3 \hat{k}$ and $b=3 \hat{i}-\hat{j}+2 \hat{k}$,then find the angle between the vectors $2 a+b$ and $a+2 b$.

$A$ vector whose modulus is $\sqrt{51}$ and makes the same angle with $a = \frac{i - 2j + 2k}{3}$,$b = \frac{-4i - 3k}{5}$,and $c = j$ is: