Let $A$ be the set of vectors $a = (a_1, a_2, a_3)$ satisfying $\left(\sum_{i=1}^3 \frac{a_i}{2^i}\right)^2 = \sum_{i=1}^3 \frac{a_i^2}{2^i}$. Then,

If the position vectors of the points $A, B, C$ are $i, j, k$ respectively and $P$ is a point such that $\overrightarrow{AB} = \overrightarrow{CP},$ then the position vector of $P$ is

If a point $C$ divides the line segment joining the points with the position vectors $2 \hat{i}-3 \hat{j}+2 \hat{k}$ and $3 \hat{i}-\hat{j}-2 \hat{k}$ in the ratio $2: 3$,then the distance of $C$ from the point with position vector $2 \hat{i}-\hat{j}+\hat{k}$ is

Represent graphically a displacement of $40 \, km,$ $30^\circ$ west of south.

If the points whose position vectors are $2 \hat{i}+\hat{j}+\hat{k}$, $6 \hat{i}-\hat{j}+2 \hat{k}$, and $14 \hat{i}-5 \hat{j}+p \hat{k}$ are collinear, then the value of $p$ is:

Find the sum of the vectors $\vec{a}=\hat{i}-2 \hat{j}+\hat{k}$,$\vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}$,and $\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}$.