If $|\vec{a}|=1, |\vec{b}|=2, |\vec{a}-\vec{b}|^2+|\vec{a}+2\vec{b}|^2=20$, then the angle $\theta$ between $\vec{a}$ and $\vec{b}$ is:

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 $\bar{r} \cdot (3\bar{i} + \bar{j} + \bar{k})$.

Let $\vec{a} = -\hat{i} - \hat{k}$,$\vec{b} = -\hat{i} + \hat{j}$,and $\vec{c} = \hat{i} + 2\hat{j} + 3\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$ and $\vec{r} \cdot \vec{a} = 0$,then the value of $\vec{r} \cdot \vec{b}$ is

Let $\vec{a}$ be a vector in the plane containing vectors $\vec{b}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=2 \hat{i}-\hat{j}+\hat{k}$. If $\vec{a}$ is perpendicular to $\hat{i}+\hat{j}+3 \hat{k}$ and its projection on $\vec{b}$ is $3 \sqrt{6}$, then $|\vec{a}|^2=$

Let $a = \sin^2 x \hat{i} + \cos^2 x \hat{j} + \hat{k}$, where $x \in R$. If the pairs of vectors $(a, \hat{i})$, $(a, \hat{j})$, and $(a, \hat{k})$ are adjacent sides of $3$ distinct parallelograms and $A$ is the sum of the squares of the areas of these parallelograms, then $A$ lies in the interval

If $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{r}$ is a non-zero vector such that $p\vec{r} + (\vec{r} \cdot \vec{b})\vec{a} = \vec{c}$,then $\vec{r} = $