If $a, b, c$ are non-coplanar vectors,then the point of intersection of the line passing through the points $2a+3b-c$ and $3a+4b-2c$ with the line joining the points $a-2b+3c$ and $a-6b+6c$ is

Let for a triangle $ABC$,
$\overline{AB} = -2\hat{i} + \hat{j} + 3\hat{k}$
$\overline{CB} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$
$\overline{CA} = 4\hat{i} + 3\hat{j} + \delta\hat{k}$
If $\delta > 0$ and the area of the triangle $ABC$ is $5\sqrt{6}$,then $\overline{CB} \cdot \overline{CA}$ is equal to

If $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$ and $|\overrightarrow{a}|=3, |\overrightarrow{b}|=4$ and $|\overrightarrow{c}|=\sqrt{37}$,then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is:

If the angle $\theta$ between the vectors $\overrightarrow{a}=2 x^2 \hat{i}+4 x \hat{j}+\hat{k}$ and $\overrightarrow{b}=7 \hat{i}-2 \hat{j}+x \hat{k}$ is such that $90^{\circ} < \theta < 180^{\circ}$,then $x$ lies in the interval

If $\bar{a}$ and $\bar{b}$ are two vectors such that $|\bar{a}|=5$,$|\bar{b}|=12$ and $|\bar{a}-\bar{b}|=13$,then $|2\bar{a}+\bar{b}|=$

If $\vec{OA} = 2\hat{i} + 2\hat{j} + \hat{k}$,$\vec{OB} = 2\hat{i} + 4\hat{j} + 4\hat{k}$ and the length of the internal bisector of $\angle BOA$ of triangle $AOB$ is $k$,then $9k^2 =$