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 $\vec{a}=4 \hat{i}-\hat{j}+\hat{k}$,$\vec{b}=11 \hat{i}-\hat{j}+\hat{k}$,and $\vec{c}$ be a vector such that $(\vec{a}+\vec{b}) \times \vec{c} = \vec{c} \times (-2 \vec{a}+3 \vec{b})$. If $(2 \vec{a}+3 \vec{b}) \cdot \vec{c} = 1670$,then $|\vec{c}|^2$ is equal to:

If $\vec{a}, \vec{b}$ and $\vec{c}$ are vectors with magnitudes $2, 3$ and $4$ respectively, then the best upper bound of $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2$ among the given values is

Let $\vec{a}=\hat{i}-2\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ be two vectors. If $\vec{c}$ is a vector such that $\vec{b} \times \vec{c}=\vec{b} \times \vec{a}$ and $\vec{c} \cdot \vec{a}=0$,then $\vec{c} \cdot \vec{b}$ is equal to

If $|\vec{a}|=2, |\vec{b}|=5$ and $|\vec{a} \times \vec{b}|=8$,then $|\vec{a} \cdot \vec{b}|$ is equal to :