For all real $x$,the vectors $Cx \hat{i} - 6 \hat{j} - 3 \hat{k}$ and $x \hat{i} + 2 \hat{j} + 2Cx \hat{k}$ make an obtuse angle with each other. Then the value of $C$ can be in:

Show that the points $A (2 \hat{i}-\hat{j}+\hat{k})$,$B (\hat{i}-3 \hat{j}-5 \hat{k})$,and $C (3 \hat{i}-4 \hat{j}-4 \hat{k})$ are the vertices of a right-angled triangle.

It is given that $a, b, c$ are vectors of lengths $6, 8, 10$ respectively. If $a$ is perpendicular to $(b+c)$, $b$ is perpendicular to $(c+a)$, and $c$ is perpendicular to $(a+b)$, then the length of the vector $a+b+c$ is (in $\sqrt{2}$)

$3 \hat{i}-2 \hat{j}-\hat{k}, -2 \hat{i}-\hat{j}+3 \hat{k}$ and $-\hat{i}+3 \hat{j}-2 \hat{k}$ are the position vectors of the vertices $A, B$ and $C$ of a $\triangle ABC$ respectively. If $H$ is its orthocenter,then $\overrightarrow{HA}+\overrightarrow{HB}+\overrightarrow{HC} = $

Suppose $\overrightarrow{a}=\lambda \hat{i}-7 \hat{j}+3 \hat{k}$ and $\overrightarrow{b}=\lambda \hat{i}+\hat{j}+2 \lambda \hat{k}$. If the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is greater than $90^{\circ}$, then $\lambda$ satisfies the inequality:

If $A = (k, 1, -1)$,$B = (2k, 0, 2)$,and $C = (2 + 2k, k, 1)$,and $AB \perp BC$,then the value of $k$ is: