If $ABCDEF$ is a regular hexagon,then $\overrightarrow {AD} + \overrightarrow {EB} + \overrightarrow {FC} = $

Vector $\vec{x}$ is a vector in the direction of $(2, -2, 1)$ and has a magnitude of $6$ units. Vector $\vec{y}$ is a vector in the direction of $(1, 1, -1)$ and has a magnitude of $\sqrt{3}$ units. Then,$|\vec{x} + 2\vec{y}| = $ . . . . . . .

If non-zero vectors $\vec{a}$,$\vec{b}$,and $\vec{c}$ are related by $\vec{a} = 8\vec{b}$ and $\vec{c} = -7\vec{b}$,find the angle between $\vec{a}$ and $\vec{c}$.

In a parallelogram $ABCD$,if $\vec{AC}$ and $\vec{BD}$ are the diagonals,then which of the following is equal to $\vec{AC} + \vec{BD}$?

Find the position vector of a point $R$ which 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.

If the position vectors of the points $A, B, C, D$ given by $\hat{i}+2 \hat{j}+3 \hat{k}, 2 \hat{i}-\hat{j}+2 \hat{k}$,$\frac{1}{4}(7 \hat{i}+15 \hat{j}+15 \hat{k})$ and $\frac{1}{3}[7 \hat{i}+2 \hat{j}+(5+3 a) \hat{k}]$ respectively are such that $|AC|=|BD|$,then $16(3a-1)^2=$