$a, b, c$ are three vectors such that $a + b + c = 0$,$|a| = 1, |b| = 2, |c| = 3$. Then $a \cdot b + b \cdot c + c \cdot a$ is equal to:

Find the angle between the vectors $2\hat{i} + 3\hat{j} + \hat{k}$ and $2\hat{i} - \hat{j} - \hat{k}$.

The vectors $a, b$ and $c$ are of the same length and taken pairwise,they form equal angles. If $a = i + j$ and $b = j + k,$ then the co-ordinates of $c$ are

$\overline{u}, \overline{v}, \overline{w}$ are three vectors such that $|\overline{u}|=1, |\overline{v}|=2, |\overline{w}|=3$. If the projection of $\overline{v}$ along $\overline{u}$ is equal to the projection of $\overline{w}$ along $\overline{u}$ and $\overline{v}, \overline{w}$ are perpendicular to each other,then $|\overline{u}-\overline{v}+\overline{w}|=$

Let $\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b}$ be a vector such that $\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$ and $\vec{a} \cdot \vec{b}=3$. Then the projection of $\vec{b}$ on the vector $\vec{a}-\vec{b}$ is :-

If $4 \hat{i}+7 \hat{j}+8 \hat{k}$,$2 \hat{i}+3 \hat{j}+4 \hat{k}$,and $2 \hat{i}+5 \hat{j}+7 \hat{k}$ are respectively the position vectors of the vertices $A, B, C$ of $\triangle ABC$,then the position vector of the point where the bisector of angle $A$ meets $BC$ is