Let $\vec{\alpha}=\hat{i}+\hat{j}+\hat{k}$, $\vec{\beta}=\hat{i}-\hat{j}-\hat{k}$ and $\vec{\gamma}=-\hat{i}+\hat{j}-\hat{k}$ be three vectors. $A$ vector $\vec{\delta}$, in the plane of $\vec{\alpha}$ and $\vec{\beta}$, whose projection on $\vec{\gamma}$ is $\frac{1}{\sqrt{3}}$, is given by

$ABC$ is a right-angled triangle in which $\max \{AB, BC, AC\} = BC$. If the position vectors of $B$ and $C$ are respectively $3\hat{i}-2\hat{j}+\hat{k}$ and $5\hat{i}+\hat{j}-3\hat{k}$,then find the value of $AB \cdot AC + BA \cdot BC + CA \cdot CB$.

Let $\vec{a}, \vec{b}, \vec{c}$ be non-coplanar vectors. If the three points with position vectors $\lambda \vec{a}-2 \vec{b}+\vec{c}$, $2 \vec{a}+\lambda \vec{b}-2 \vec{c}$, and $4 \vec{a}+7 \vec{b}-8 \vec{c}$ are collinear, then $\lambda=$

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\overrightarrow{b} = \hat{i} - \hat{j} + 2\hat{k}$,and $\overrightarrow{c} = 5\hat{i} - 3\hat{j} + 3\hat{k}$ be three vectors. If $\overrightarrow{r}$ is a vector such that $\overrightarrow{r} \times \overrightarrow{b} = \overrightarrow{c} \times \overrightarrow{b}$ and $\overrightarrow{r} \cdot \overrightarrow{a} = 0$,then $25|\overrightarrow{r}|^2$ is equal to:

$A$ vector $\overrightarrow{a} = \alpha \hat{i} + 2 \hat{j} + \beta \hat{k}$ (where $\alpha, \beta \in R$) lies in the plane of the vectors $\overrightarrow{b} = \hat{i} + \hat{j}$ and $\overrightarrow{c} = \hat{i} - \hat{j} + 4 \hat{k}$. If $\overrightarrow{a}$ bisects the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$,then:

If $\vec{a} + \vec{b} + \vec{c} = \vec{0}$,$|\vec{a}| = 3$,$|\vec{b}| = 5$,and $|\vec{c}| = 7$,find the angle between $\vec{a}$ and $\vec{b}$.