The value of $c$ such that for all real $x$, the vectors $\vec{a} = cxi - 6j + 3k$ and $\vec{b} = xi + 2j + 2cxk$ make an obtuse angle is:

In a triangle $ABC$, $D$ and $E$ divide the sides $BC$ and $CA$ in the ratio $2:1$ respectively. If $P$ is the point of intersection of $AD$ and $BE$, then the ratio in which $P$ divides $AD$ is

If $\overrightarrow{F_1} = i - j + k,$ $\overrightarrow{F_2} = -i + 2j - k,$ $\overrightarrow{F_3} = j - k,$ $\vec{A} = 4i - 3j - 2k$ and $\vec{B} = 6i + j - 3k,$ then the scalar product of $(\overrightarrow{F_1} + \overrightarrow{F_2} + \overrightarrow{F_3})$ and $\overrightarrow{AB}$ will be:

If the vectors $\vec{BC} = 2\hat{i} + \hat{j} + \hat{k}$ and $\vec{CD} = \hat{i} + 2\hat{j} - 2\hat{k}$ represent two adjacent sides of a parallelogram $ABCD$ and $\theta$ is the angle between its diagonals $\vec{AC}$ and $\vec{BD}$, then $\tan \theta =$

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be vectors of equal magnitude such that the angle between $\vec{a}$ and $\vec{b}$ is $\alpha$, $\vec{b}$ and $\vec{c}$ is $\beta$, and $\vec{c}$ and $\vec{a}$ is $\gamma$. Then the minimum value of $\cos \alpha + \cos \beta + \cos \gamma$ is

Let $\overrightarrow{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$ and $\overrightarrow{b} = 7\hat{i} + \hat{j} - 6\hat{k}$. If $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{r} \times \overrightarrow{b}$ and $\overrightarrow{r} \cdot (\hat{i} + 2\hat{j} + \hat{k}) = -3$,then $\overrightarrow{r} \cdot (2\hat{i} - 3\hat{j} + \hat{k})$ is equal to: