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

Let $\lambda \in \mathbb{Z}$,$\vec{a} = \lambda \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 3 \hat{i} - \hat{j} + 2 \hat{k}$. Let $\vec{c}$ be a vector such that $(\vec{a} + \vec{b} + \vec{c}) \times \vec{c} = \vec{0}$,$\vec{a} \cdot \vec{c} = -17$ and $\vec{b} \cdot \vec{c} = -20$. Then $|\vec{c} \times (\lambda \hat{i} + \hat{j} + \hat{k})|^2$ is equal to

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{c} = \hat{j} - \hat{k}$ and a vector $\vec{b}$ be such that $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{a} \cdot \vec{b} = 3$. Then $|\vec{b}|$ equals?

$A$ unit vector in the $xy$-plane which is perpendicular to $4i - 3j + k$ is

Let a unit vector which makes an angle of $60^{\circ}$ with $2 \hat{i}+2 \hat{j}-\hat{k}$ and an angle of $45^{\circ}$ with $\hat{i}-\hat{k}$ be $\overrightarrow{C}$. Then $\overrightarrow{C}+\left(-\frac{1}{2} \hat{i}+\frac{1}{3 \sqrt{2}} \hat{j}-\frac{\sqrt{2}}{3} \hat{k}\right)$ is :

Let $\overrightarrow{a} = 2\hat{i} + \hat{j} + \hat{k}$,and $\overrightarrow{b}$ and $\overrightarrow{c}$ be two nonzero vectors such that $|\vec{a} + \vec{b} + \vec{c}| = |\vec{a} + \vec{b} - \vec{c}|$ and $\vec{b} \cdot \vec{c} = 0$. Consider the following two statements:
$(A)$ $|\overrightarrow{a} + \lambda \overrightarrow{c}| \geq |\overrightarrow{a}|$ for all $\lambda \in R$.
$(B)$ $\overrightarrow{a}$ and $\overrightarrow{c}$ are always parallel.