If $\bar{a}, \bar{b}, \bar{c}$ are three unit vectors such that $|\bar{a}-\bar{b}|^2+|\bar{b}-\bar{c}|^2+|\bar{c}-\bar{a}|^2=15$,then $|\bar{a}-\bar{b}-\bar{c}|^2-4(\bar{b} \cdot \bar{c})=$

The angle between two vectors $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ is . . . . . . .

If the vectors $\vec{a} = \hat{i} - 2x\hat{j} - 3y\hat{k}$ and $\vec{b} = \hat{i} + 3x\hat{j} + 2y\hat{k}$ are perpendicular to each other,find the locus of the point $(x, y)$.

Let $\vec{a}=6 \hat{i}+9 \hat{j}+12 \hat{k}$,$\vec{b}=\alpha \hat{i}+11 \hat{j}-2 \hat{k}$ and $\vec{c}$ be vectors such that $\vec{a} \times \vec{c}=\vec{a} \times \vec{b}$. If $\vec{a} \cdot \vec{c}=-12$ and $\vec{c} \cdot (\hat{i}-2 \hat{j}+\hat{k})=5$,then $\vec{c} \cdot (\hat{i}+\hat{j}+\hat{k})$ is equal to $.............$.

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 the position vectors of the vertices $A$,$B$ and $C$ respectively of triangle $ABC$,then the position vector of the point in which the bisector of $\angle B$ meets $CA$ is:

If $\bar{a}=\hat{i}+2 \hat{j}+\hat{k}$,$\bar{b}=\hat{i}-\hat{j}+\hat{k}$,and $\bar{c}=\hat{i}+\hat{j}-\hat{k}$,then a vector in the plane of $\bar{a}$ and $\bar{b}$,whose projection on $\bar{c}$ is $\frac{1}{\sqrt{3}}$,is