If the vectors $\overline{a} = c(\log_7 x) \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $\overline{b} = (\log_7 x) \hat{i} + 3c(\log_7 x) \hat{j} - 4 \hat{k}$ make an obtuse angle for any $x > 0$,then $c$ belongs to

If the vectors $a\hat{i} + 2\hat{j} + 3\hat{k}$ and $-\hat{i} + 5\hat{j} + a\hat{k}$ are perpendicular to each other,then $a = $

$\bar{a}, \bar{b}, \bar{c}$ are nonzero vectors such that $\bar{a}$ is perpendicular to $\bar{b}$ and $\bar{c}$,$|\bar{a}|=1, |\bar{b}|=2, |\bar{c}|=1$ and $\bar{b} \cdot \bar{c}=1$. There is a nonzero vector $\bar{d}$ coplanar with $\bar{a}+\bar{b}$ and $2\bar{b}-\bar{c}$. If $\bar{d} \cdot \bar{a}=1$,then $|\bar{d}|^2=$ (Note that $x$ and $y$ are parameters involved when we write $\bar{d}=x(\bar{a}+\bar{b})+y(2\bar{b}-\bar{c})$)

$A$ particle is acted upon by constant forces $4\hat{i} + \hat{j} - 3\hat{k}$ and $3\hat{i} + \hat{j} - \hat{k}$. The displacement of the particle from the point $\hat{i} + 2\hat{j} + 3\hat{k}$ to the point $5\hat{i} + 4\hat{j} + \hat{k}$ is given. Find the total work done by the forces in units.

If $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$,$|\overrightarrow{a}|=3$,$|\overrightarrow{b}|=5$,and $|\overrightarrow{c}|=7$,then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is

$P$ is the circumcentre of $\triangle ABC$. If the position vectors of $A, B, C$ and $P$ are $\bar{a}, \bar{b}, \bar{c}$ and $\frac{\bar{a}+\bar{b}+\bar{c}}{4}$ respectively,then the position vector of the orthocentre of this triangle is