Let $\vec{b}$ and $\vec{c}$ be non-collinear vectors satisfying $\vec{a} \times (\vec{b} \times \vec{c}) + (\vec{a} \cdot \vec{b})\vec{b} = (4 - 2x - \sin y)\vec{b} + (x^2 - 1)\vec{c}$ and $(\vec{c} \cdot \vec{c})\vec{a} = \vec{c}$,then $x$ is equal to

If $P=(0,1,2), Q=(4,-2,1)$ and $O=(0,0,0)$,then $\angle POQ=$

If $|\bar{a}|=2, |\bar{b}|=3$ and $\bar{a}, \bar{b}$ are mutually perpendicular vectors,then the area of the triangle whose vertices are $0, \bar{a}+2\bar{b}, \bar{a}-2\bar{b}$ 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

$AB=a$ and $AC=b$ are the sides of $\triangle ABC$. $P$ is a point on $AB$ and $Q$ is a point on $BC$ such that $\frac{AP}{PB}=\frac{1}{2}$ and $\frac{BQ}{QC}=\frac{1}{2}$. If the point of intersection of $AQ$ and $CP$ is $D$ and the area of $\triangle BCD$ is $7$ square units,then the area of the $\triangle ABC$ (in the same square units) is

Let $\hat{a}$ and $\hat{b}$ be unit vectors. If $\vec{c}$ is a vector such that the angle between $\hat{a}$ and $\vec{c}$ is $\frac{\pi}{12}$,and $\hat{b} = \vec{c} + 2(\vec{c} \times \hat{a})$,then $|6\vec{c}|^{2}$ is equal to