In $\triangle ABC$,if $D$ and $E$ are the mid-points of the sides $BC$ and $CA$ respectively,then $2(\vec{AD}+\vec{EB})=$

If the vector $a = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ and $b$ are collinear and $|b| = 21$, then $b$ is equal to:

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=\hat{i}-2 \hat{j}+\hat{k},$ find a unit vector parallel to the vector $2 \vec{a}-\vec{b}+3 \vec{c}.$

If $\bar{a}, \bar{b}, \bar{c}$ are non-zero vectors such that no two of them are parallel,and $\bar{a} + \bar{b}$ is parallel to $\bar{c}$,and $\bar{b} + \bar{c}$ is parallel to $\bar{a}$,then $\bar{a} + \bar{b} + \bar{c} = $

$R$ divides the line joining two points $P$ and $Q$ whose position vectors are $\hat{i}+2 \hat{j}-\hat{k}$ and $-\hat{i}+\hat{j}+\hat{k}$ respectively in the ratio $2: 1$ externally. $S$ divides $PQ$ internally in the ratio $2: 1$. Then,the position vector of the midpoint of the line joining $R$ and $S$ is

In triangle $ABC$ (Fig.),which of the following is not true: