If the position vectors of the points $A$ and $B$ are $\vec{a} = \hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b} = 3\hat{i} - \hat{j} - 3\hat{k}$,then what will be the position vector of the midpoint of $AB$?

The position vectors of two points $A$ and $B$ are $\bar{i}+2\bar{j}+3\bar{k}$ and $7\bar{i}-\bar{k}$ respectively. The point $P$ with position vector $-2\bar{i}+3\bar{j}+5\bar{k}$ is on the line $AB$. If the point $Q$ is the harmonic conjugate of $P$ with respect to $A$ and $B$,then the sum of the scalar components of the position vector of $Q$ is

Vectors $\vec{a}$,$\vec{b}$,and $\vec{c}$ are of the same length and they make equal angles with each other when taken in pairs. If $\vec{a} = \hat{i} - \hat{j}$,$\vec{b} = \hat{j} + \hat{k}$,and $\vec{c}$ makes an obtuse angle with the $x$-axis,find the vector $\vec{c}$.

Let the three sides of a triangle $ABC$ be represented by the vectors $\vec{AB} = 2\hat{i}-\hat{j}+\hat{k}$,$\vec{BC} = 3\hat{i}-4\hat{j}-4\hat{k}$,and $\vec{CA} = \hat{i}-3\hat{j}-5\hat{k}$. Let $G$ be the centroid of the triangle $ABC$. Then $6(|\overrightarrow{AG}|^2+|\overrightarrow{BG}|^2+|\overrightarrow{CG}|^2)$ is equal to

If $\theta$ is the interior angle of a regular pentagon,then $|(\sin \theta) \hat{i}+(\cos \theta) \hat{j}+(\tan \theta) \hat{k}|=$

For any vector $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$,with $10|a_i| < 1$,$i = 1, 2, 3$,consider the following statements:
$(A): \max \{|a_1|, |a_2|, |a_3|\} \leq |\vec{a}|$
$(B): |\vec{a}| \leq 3 \max \{|a_1|, |a_2|, |a_3|\}$