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

The projection of the vector $2i + j - 3k$ on the vector $i - 2j + k$ is:

Magnitudes of vectors $\vec{a}, \vec{b}, \vec{c}$ are $3, 4, 5$ respectively. If $\vec{a}$ and $\vec{b} + \vec{c}$,$\vec{b}$ and $\vec{c} + \vec{a}$,and $\vec{c}$ and $\vec{a} + \vec{b}$ are mutually perpendicular,then the magnitude of $\vec{a} + \vec{b} + \vec{c}$ 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:

If three unit vectors $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ satisfy $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$,then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is

If $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$,$\vec{b}=-\hat{i}+2\hat{j}+\hat{k}$,$\vec{c}=3\hat{i}+\hat{j}$ and $\vec{a}+\lambda\vec{b}$ is perpendicular to $\vec{c}$,then $\lambda=$