If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}+3\hat{k}$,then $(\vec{a}+\vec{b}) \cdot (\vec{a}-\vec{b}) = $ . . . . . . .

Let $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ be the position vector of a point $A$. Let $\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$ and $\vec{c}=\hat{i}+\hat{j}-2 \hat{k}$ be two vectors and $\vec{r}$ be a vector passing through the point $A$ with position vector $\vec{a}$ and parallel to the vector $\vec{b}$. If the projection of $\vec{r}$ on $\vec{c}$ is $\frac{9}{\sqrt{6}}$,then find $|\vec{r}|$.

Let $\theta$ denote the angle between vectors $\vec{a}$ and $\vec{b}$. If $\vec{a}=2 \hat{i}+3 \hat{j}+6 \hat{k}$,$\vec{a} \cdot \vec{b}=4$ and $\theta=\cos ^{-1}\left(\frac{4}{21}\right)$,then $\vec{a}+\vec{b}$ is:

The vectors $\vec{AB} = 3\hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{BC} = \hat{i} - 2\hat{k}$ are the adjacent sides of a parallelogram. The angle between its diagonals is

If $\theta$ is the angle between the unit vectors $a$ and $b$,then $\sin \frac{\theta}{2}$ is equal to

If the moduli of $a$ and $b$ are equal and the angle between them is $120^\circ$ and $a \cdot b = -8$,then $|a|$ is equal to