If $\bar{a}, \bar{b}, \bar{c}$ are three non-zero vectors,no two of them are collinear,$\bar{a}+2 \bar{b}$ is collinear with $\bar{c}$,and $\bar{b}+3 \bar{c}$ is collinear with $\bar{a}$,then $\bar{a}+2 \bar{b}$ is equal to:

The ratio in which $\hat{i}+2 \hat{j}+3 \hat{k}$ divides the join of $-2 \hat{i}+3 \hat{j}+5 \hat{k}$ and $7 \hat{i}-\hat{k}$ is

If the position vectors of the vertices $A, B, C, D$ of a quadrilateral are $7 \hat{i}-4 \hat{j}+7 \hat{k}, \hat{i}-6 \hat{j}+10 \hat{k}, -\hat{i}-3 \hat{j}+4 \hat{k}$,and $5 \hat{i}-\hat{j}+5 \hat{k}$ respectively,then $ABCD$ is

$D, E, F$ are points on the sides $BC, CA$ and $AB$ of a $\triangle ABC$ respectively,dividing them in the ratios $2:3, 1:2, 3:1$ internally. The lines $BE$ and $CF$ intersect on the line $AD$ at $P$. If $\overrightarrow{AP} = x_1 \overrightarrow{AB} + y_1 \overrightarrow{AC}$,then $x_1 + y_1 =$

In $\triangle PQR$,$(4 \hat{i}+3 \hat{j}+6 \hat{k})$,$(2 \hat{i}+2 \hat{j}+3 \hat{k})$ and $(3 \hat{i}+\hat{j}+3 \hat{k})$ are the position vectors of the vertices $P, Q$ and $R$ respectively. Then the position vector of the point of intersection of the angle bisector of $P$ with $QR$ is

$R$ 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