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

Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $(2 \vec{a}+\vec{b})$ and $(\vec{a}-3 \vec{b})$ externally in the ratio $1:2$. Also,show that $P$ is the mid-point of the line segment $RQ$.

In a $\triangle ABC$ (shown in the figure below),state whether the following are true or false:
$(i)$ $\vec{AB} + \vec{BC} + \vec{CA} = \vec{0}$
(ii) $\vec{AB} + \vec{BC} - \vec{AC} = \vec{0}$
(iii) $\vec{AB} - \vec{CB} + \vec{CA} = \vec{0}$
(iv) $\vec{AB} + \vec{BC} - \vec{CA} = \vec{0}$

Let $\overrightarrow{OA} = \hat{i} - 3\hat{j} + \hat{k}$, $\overrightarrow{OB} = \hat{i} + 3\hat{j} - 2\hat{k}$, and $\overrightarrow{OC} = 4\hat{i} + 3\hat{j} + 5\hat{k}$ be the position vectors of three points $A$, $B$, and $C$. Let $P$ be the point which divides $AB$ in the ratio $2:1$. If $l, m, n$ are the direction cosines of the vector $\overrightarrow{PC}$, then $l + 3m + 2n =$

The points in space represented by the position vectors $A = 4\hat{i}+\hat{j}+3\hat{k}$,$B = 6\hat{i}-2\hat{j}-3\hat{k}$,and $C = \hat{i}-\hat{j}-3\hat{k}$ form:

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