Evaluate the product $(3 \vec{a}-5 \vec{b}) \cdot (2 \vec{a}+7 \vec{b})$.

Let $\overrightarrow{a} = 2\hat{i} + \hat{j} + \hat{k}$,and $\overrightarrow{b}$ and $\overrightarrow{c}$ be two nonzero vectors such that $|\vec{a} + \vec{b} + \vec{c}| = |\vec{a} + \vec{b} - \vec{c}|$ and $\vec{b} \cdot \vec{c} = 0$. Consider the following two statements:
$(A)$ $|\overrightarrow{a} + \lambda \overrightarrow{c}| \geq |\overrightarrow{a}|$ for all $\lambda \in R$.
$(B)$ $\overrightarrow{a}$ and $\overrightarrow{c}$ are always parallel.

If the position vectors of the vertices $A, B, C$ of a triangle $ABC$ are $4 \hat{\imath} + 7 \hat{\jmath} + 8 \hat{k}$,$2 \hat{\imath} + 3 \hat{\jmath} + 4 \hat{k}$,and $2 \hat{\imath} + 5 \hat{\jmath} + 7 \hat{k}$ respectively,then the position vector of the point where the bisector of angle $A$ meets $BC$ is

Let $\vec{a} = \sqrt{7}\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{k}$. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}$ and $\vec{r} \cdot \vec{a} = 0$, then $|3\vec{r}|^2$ is equal to:

Let the arc $AC$ of a circle subtend a right angle at the centre $O$. If the point $B$ on the arc $AC$ divides the arc $AC$ such that $\frac{\text{length of arc } AB}{\text{length of arc } BC} = \frac{1}{5}$,and $\overrightarrow{OC} = \alpha \overrightarrow{OA} + \beta \overrightarrow{OB}$,then $\alpha + \sqrt{2}(\sqrt{3}-1) \beta$ is equal to

Let the points $A, B$ and $P$ be $(-2, 2, 4), (2, 6, 3)$ and $(1, 2, 1)$ respectively. The magnitude of the moment of the force represented by $\overrightarrow{AB}$ and acting at $A$ about $P$ is