If $a = x^2 \hat{i} + x \hat{j} + 3 \hat{k}$ and $b = x \hat{i} - 4 \hat{j} + 2 \hat{k}$ and $a \cdot b > 6$,then:

Three vectors $\vec a, \vec b, \vec c$ are inclined at an acute angle with each other such that $|\vec a| = 2, |\vec b| = 3, |\vec c| = 9$ and the lengths of the projections of $\vec a$ on $\vec b$,$\vec b$ on $\vec c$,and $\vec c$ on $\vec a$ respectively are in geometric progression. If the angle between $\vec a$ and $\vec b$ is $\frac{5\pi}{12}$ and the angle between $\vec c$ and $\vec a$ is $\frac{\pi}{12}$,then the angle between $\vec b$ and $\vec c$ is:

If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors such that $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{c}$,and $|\vec{a} + \vec{b} + \vec{c}| = 1$,then the angle between $\vec{b}$ and $\vec{c}$ is:

$a, b, c$ are three vectors such that $|a|=3, |b|=5, |c|=7$. If $a, b, c$ are perpendicular to the vectors $b+c, c+a, a+b$ respectively,then $\sqrt{(a+b+c)^2-2}=$

$A$ parallelogram is constructed with $5\vec{a} + 2\vec{b}$ and $\vec{a} - 3\vec{b}$ as its adjacent sides, where $|\vec{a}| = 2\sqrt{2}$ and $|\vec{b}| = 3$. The angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$. Find the lengths of the diagonals of the parallelogram.

Lines $\vec{r} = \vec{a} + \lambda\vec{b}$ and $\vec{r} = \vec{b} + \mu\vec{a}$ intersect at point $(2, 4, -4)$. If $|\vec{a} - \vec{b}| = 4$, then $\vec{a} \cdot \vec{b} =$