The set of all real values of $c$ such that the angle between the vectors $\vec{a} = cx \hat{i} - 6 \hat{j} + 3 \hat{k}$ and $\vec{b} = x \hat{i} + 2 \hat{j} + 2cx \hat{k}$ is an obtuse angle for all real $x$ is:

Let three vectors $\overrightarrow{a}=\alpha \hat{i}+4 \hat{j}+2 \hat{k}$,$\vec{b}=5 \hat{i}+3 \hat{j}+4 \hat{k}$,and $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$ form a triangle such that $\overrightarrow{c}=\overrightarrow{a}-\overrightarrow{b}$ and the area of the triangle is $5 \sqrt{6}$. If $\alpha$ is a positive real number,then $|\overrightarrow{c}|^2$ is:

For any two non-zero vectors $\vec{a}$ and $\vec{b}$,$(a \vec{b} + b \vec{a}) \cdot (a \vec{b} - b \vec{a})$ is equal to:

If $a, b, c$ are unit vectors such that $a + b + c = 0,$ then $a \cdot b + b \cdot c + c \cdot a = $

Let $\bar{a} = \bar{i} + 2\bar{j} + 3\bar{k}$, $\bar{b} = 2\bar{i} - 3\bar{j} + \bar{k}$, and $\bar{c} = 3\bar{i} + \bar{j} - 2\bar{k}$ be three vectors. If $\bar{r}$ is a vector such that $\bar{r} \cdot \bar{a} = 0$, $\bar{r} \cdot \bar{b} = -2$, and $\bar{r} \cdot \bar{c} = 6$, then find the value of $\bar{r} \cdot (3\bar{i} + \bar{j} + \bar{k})$.

If $a$ is a unit vector, then $|a \times \hat{i}|^2+|a \times \hat{j}|^2+|a \times \hat{k}|^2=$