Position of a particle in a rectangle co-ordinate system is $(3,2,5)$. Then its position vector will be

Q. The vector $3\hat{i}+5\hat{j}+2\hat{k},2\hat{i}-3\hat{j}-5\hat{k}$ and $5\hat{i}+2\hat{j}-3\hat{k}$ form the sides of a/an

Q. Following forces start acting on a particle at rest at the origin of the co-ordinate system simultaneously
${\overrightarrow{F}}_1=-4\hat{i}-5\hat{j}+5\hat{k}$,${\overrightarrow{F}}_2=5\hat{i}+8\hat{j}=6\hat{k},$${\overrightarrow{F}}_3=-3\hat{i}+4\hat{j}-7\hat{k}$$\text{and}{\overrightarrow{F}}_4=2\hat{i}-3\hat{j}-2\hat{k}$

Q. Following forces start acting on a particle at rest at the origin of the co-ordinate system simultaneously $\overrightarrow{F_1}=-4\hat{i}-5\hat{j}+5\hat{k},\overrightarrow{F_2}=-5\hat{i}+8\hat{j}+6\hat{k},\overrightarrow{F_3}=-3\hat{i}+4\hat{j}-7\hat{k}$ and $\overrightarrow{F_4}=12\hat{i}-3\hat{j}-2\hat{k}$ then the particle will move

Q. Two forces $P=3\hat{i}-2\hat{j}+\hat{k}$ and $Q=\hat{i}+3\hat{j}-5\hat{k}$ acting on a particle $A$ move it to $B$. If the position vectors of $A$ and $B$ are respectively $-2\hat{i}+5\hat{k}$ and $3\hat{i}-7\hat{j}+2\hat{k}$, then the total work done is

Q. If the position vectors of $A,B,C,D$ are $3\hat{i}+2\hat{j}+\hat{k},4\hat{i}+5\hat{j}+5\hat{k},4\hat{i}+2\hat{j}-2\hat{k},6\hat{i}+5\hat{j}-\hat{k}$ respectively then the position vector of the point of intersection of $\overline{AB}$ and $\overline{CD}$ is