In the triangle $ABC,$ if $\overrightarrow{AB} = a, \overrightarrow{AC} = c, \overrightarrow{BC} = b$,then which of the following is correct?

If $\overrightarrow{AB} = (2\hat{i} + 3\hat{j} - \hat{k})$ and $A(1, 2, -1)$ is the given point,then the coordinates of $B$ are

If $\vec{a} = 7\hat{j} + 10\hat{k}$, $\vec{b} = -\hat{i} + 6\hat{j} + 6\hat{k}$ and $\vec{c} = -4\hat{i} + 9\hat{j} + 6\hat{k}$ are the position vectors of the vertices $A, B$ and $C$ respectively of $\triangle ABC$. Then the position vector of the point where the bisector of the angle $A$ meets side $BC$ is

Three non-zero non-collinear vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are such that $\vec{a}+3\vec{b}$ is collinear with $\vec{c}$, and $3\vec{b}+2\vec{c}$ is collinear with $\vec{a}$. Then $\vec{a}+3\vec{b}+2\vec{c}$ equals to

If $\vec{a}=(2x+y)\hat{i}+3\hat{j}+9\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}-(x-y)\hat{k}$ are two collinear vectors,then $x^3+27y^3=$

The triad $(x, y, z)$ of real numbers such that $(3 \hat{i}-\hat{j}+2 \hat{k})=(2 \hat{i}+3 \hat{j}-\hat{k}) x+(\hat{i}-2 \hat{j}+2 \hat{k}) y+(-2 \hat{i}+\hat{j}-2 \hat{k}) z$ is