Vector $\vec{a} + 3\vec{b}$ is perpendicular to $7\vec{a} - 5\vec{b}$ and $\vec{a} - 5\vec{b}$ is perpendicular to $7\vec{a} + 3\vec{b}$. The angle between non-zero vectors $\vec{a}$ and $\vec{b}$ is:

The work done by the force $F = 2i - 3j + 2k$ in displacing a particle from the point $(3, 4, 5)$ to the point $(1, 2, 3)$ is ............ $unit$.

The angle between two consecutive sides $\vec{a}$ and $\vec{b}$ of a parallelogram is $\frac{\pi}{6}$. Given $\vec{a} = (2, -2, 1)$ and $\vec{b} = 2|\vec{a}|$,the area of the parallelogram is . . . . . . sq. units.

Consider the following Assertion $(A)$ and Reason $(R)$:
Assertion $(A)$: The two lines $\bar{r}=\bar{a}+t(\bar{b})$ and $\bar{r}=\bar{b}+s(\bar{a})$ intersect each other.
Reason $(R)$: The shortest distance between the lines $\bar{r}=\bar{p}+t(\bar{q})$ and $\bar{r}=\bar{c}+s(\bar{d})$ is equal to the length of the projection of the vector $(\bar{p}-\bar{c})$ on $(\bar{q} \times \bar{d})$.
The correct answer is:

If with reference to the right-handed system of mutually perpendicular unit vectors $\hat{i}, \hat{j}$ and $\hat{k}$,$\vec{\alpha} = 3\hat{i} - \hat{j}$ and $\vec{\beta} = 2\hat{i} + \hat{j} - 3\hat{k}$,then express $\vec{\beta}$ in the form $\vec{\beta} = \vec{\beta}_{1} + \vec{\beta}_{2}$,where $\vec{\beta}_{1}$ is parallel to $\vec{\alpha}$ and $\vec{\beta}_{2}$ is perpendicular to $\vec{\alpha}$.

The value of $|\vec{a} \cdot \vec{b}|^2 + |\vec{a} \times \vec{b}|^2$ is...