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

Let $\overline{u}, \overline{v}, \overline{w}$ be vectors such that $|\overline{u}|=1, |\overline{v}|=2, |\overline{w}|=3$. If the projection of $\overline{v}$ along $\overline{u}$ is equal to that of $\overline{w}$ along $\overline{u}$,and the vectors $\overline{v}$ and $\overline{w}$ are perpendicular to each other,then $|\overline{u}-\overline{v}+\overline{w}|$ equals:

If $\vec{a}=\hat{i}+\lambda \hat{j}+2 \hat{k}$ and $\vec{b}=\mu \hat{i}+\hat{j}-\hat{k}$ are orthogonal and $|\vec{a}|=|\vec{b}|$,then $(\lambda, \mu) = $

Let $|\vec{a}| = |\vec{b}| = 1$ and $|\vec{a} + \vec{b}| = \sqrt{3}$. If $\vec{c}$ is a vector satisfying the condition $\vec{c} - \vec{a} - 2\vec{b} = 3(\vec{a} \times \vec{b})$,then $\vec{c} \cdot \vec{b} = \dots$ (in $/2$)

If $\overrightarrow{a}=-\hat{i}+\hat{j}+2 \hat{k}$, $\overrightarrow{b}=2 \hat{i}-\hat{j}-\hat{k}$ and $\overrightarrow{c}=-2 \hat{i}+\hat{j}+3 \hat{k}$, then the angle between $2 \overrightarrow{a}-\overrightarrow{c}$ and $\overrightarrow{a}+\overrightarrow{b}$ is

If the vectors $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$,$\vec{b} = 2\hat{i} + 4\hat{j} + \hat{k}$,and $\vec{c} = \alpha\hat{i} + \hat{j} + \beta\hat{k}$ are mutually orthogonal,then $(\alpha, \beta) = $