If $\triangle ABC$ is right-angled at $A$,where $A \equiv (4, 2, x)$,$B \equiv (3, 1, 8)$,and $C \equiv (2, -1, 2)$,then the value of $x$ is

Let $\overrightarrow{a}=\hat{i}+\alpha \hat{j}+\beta \hat{k}$,where $\alpha, \beta \in R$. Let a vector $\overrightarrow{b}$ be such that the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$ and $|\vec{b}|^2=6$. If $\vec{a} \cdot \vec{b}=3 \sqrt{2}$,then the value of $(\alpha^2+\beta^2)|\vec{a} \times \vec{b}|^2$ is equal to

Let $\overrightarrow{OA}$ and $\overrightarrow{OB}$ be two sides of a triangle. The median $\overrightarrow{AM}$ is perpendicular to the angle bisector $\overrightarrow{OL}$ and $|\overrightarrow{AM}|:|\overrightarrow{OL}|=1:2$. The angle between $\overrightarrow{OA}$ and $\overrightarrow{OB}$ is

If $a$ makes an acute angle with $b$,$r \cdot a = 0$ and $r \times b = c \times b$,then $r=$

Let $\vec{AB} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $\vec{AD} = \hat{i} + 2 \hat{j} + \lambda \hat{k}$, $\lambda \in R$. Let the projection of the vector $\vec{v} = \hat{i} + \hat{j} + \hat{k}$ on the diagonal $\vec{AC}$ of the parallelogram $ABCD$ be of length $1$ unit. If $\alpha, \beta$, where $\alpha > \beta$, are the roots of the equation $\lambda^2 x^2 - 6 \lambda x + 5 = 0$, then $2 \alpha - \beta$ is equal to

Find the distance of the point $P(-\hat{i} + 2\hat{j} + 6\hat{k})$ from the line passing through the point $A(2\hat{i} + 3\hat{j} - 4\hat{k})$ and parallel to the vector $\vec{b} = 6\hat{i} + 3\hat{j} - 4\hat{k}$.