If $A=(1,-1,2)$, $B=(3,4,-2)$, $C=(0,3,2)$ and $D=(3,5,6)$, then the angle between the lines $\overrightarrow{AB}$ and $\overrightarrow{CD}$ is (in $^{\circ}$)

$\bar{a}, \bar{b}, \bar{c}$ are nonzero vectors such that $\bar{a}$ is perpendicular to $\bar{b}$ and $\bar{c}$,$|\bar{a}|=1, |\bar{b}|=2, |\bar{c}|=1$ and $\bar{b} \cdot \bar{c}=1$. There is a nonzero vector $\bar{d}$ coplanar with $\bar{a}+\bar{b}$ and $2\bar{b}-\bar{c}$. If $\bar{d} \cdot \bar{a}=1$,then $|\bar{d}|^2=$ (Note that $x$ and $y$ are parameters involved when we write $\bar{d}=x(\bar{a}+\bar{b})+y(2\bar{b}-\bar{c})$)

If $a, b, c$ are non-zero vectors such that $a \cdot b = a \cdot c$,then which statement is true?

Two adjacent sides of a triangle are represented by the vectors $\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{b} = 2\sqrt{3}\hat{i} - 2\sqrt{3}\hat{j} + \sqrt{3}\hat{k}$. Then the least angle of the triangle and the perimeter of the triangle are respectively:

If for vectors $\vec{a}, \vec{b},$ and $\vec{c}$,$\vec{a} + \vec{b} + \vec{c} = \vec{0}$ and $|\vec{a}| = 7, |\vec{b}| = 5, |\vec{c}| = 3$,then the angle between $\vec{b}$ and $\vec{c}$ is ............ $^o$.

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{c}$, and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2 \pi}{3}$, then $|\vec{a}+3 \vec{b}-4 \vec{c}|^2=$