Three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ satisfy $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. If $|\vec{a}|=3, |\vec{b}|=4, |\vec{c}|=2$, then $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}+2(|\vec{a}|+|\vec{b}|+|\vec{c}|)=$

The orthogonal projection of vector $a$ on vector $b$ is given by:

If vectors satisfy the condition $|a - c| = |b - c|$,then $(b - a) \cdot \left( c - \frac{a + b}{2} \right)$ is equal to

Let $\bar{a} = 2\bar{i} - \bar{j} + \bar{k}$,$\bar{b} = \bar{i} + 2\bar{j} - \bar{k}$,and $\bar{c} = \bar{i} + \bar{j} - 2\bar{k}$ be three vectors. $A$ vector $\bar{r}$ in the plane of $\bar{b}$ and $\bar{c}$ has a projection of magnitude $\sqrt{\frac{2}{3}}$ on the vector $\bar{a}$. Find $\bar{r}$.

In a parallelogram $OACB$,$\overrightarrow{OA} = \vec{a}$,$\overrightarrow{OB} = \vec{b}$,and the foot of the perpendicular drawn from point $B$ to $AC$ is $M$. If $\vec{a} \cdot \vec{b} = 1$ and $|\vec{a}| = |\vec{b}| = 2$,then $|\overrightarrow{BM}|$ is:

The projection of the vector $\hat{i} + \hat{j} + \hat{k}$ on the line $\vec{r} = 3\hat{i} - \hat{j} + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})$ is: