If $|\vec{a}| = \sqrt{27}$,$|\vec{b}| = 7$ and $|\vec{a} \times \vec{b}| = 35$,then $\vec{a} \cdot \vec{b}$ is equal to

If $(2, -1, 2)$ and $(K, 3, 5)$ are the triads of direction ratios of two lines and the angle between them is $45^{\circ}$,then the value of $K$ is

If the force $\overrightarrow{F} = \hat{i} + 2\hat{j} + 3\hat{k}$ moves a particle from position $\vec{r_1} = \hat{i} + \hat{j} - \hat{k}$ to $\vec{r_2} = 2\hat{i} - \hat{j} + \hat{k},$ then the work done is:

Let a vector $\overrightarrow{a}=\sqrt{2}\hat{i}-\hat{j}+\lambda\hat{k}$, $\lambda>0$, make an obtuse angle with the vector $\overrightarrow{b}=-\lambda^{2}\hat{i}+4\sqrt{2}\hat{j}+4\sqrt{2}\hat{k}$ and an angle $\theta$, $\frac{\pi}{6} < \theta < \frac{\pi}{2}$, with the positive $z$-axis. If the set of all possible values of $\lambda$ is $(\alpha, \beta)-\{\gamma\}$, then $\alpha+\beta+\gamma$ is equal to . . . . . . .

If $p$-th, $q$-th, and $r$-th terms of a geometric progression are the positive numbers $a, b,$ and $c$ respectively, then the angle between the vectors $(\log a^2) i + (\log b^2) j + (\log c^2) k$ and $(q-r) i + (r-p) j + (p-q) k$ is

The angle between two vectors $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ is . . . . . . .