The values of $a$,for which points $A, B, C$ with position vectors $2\hat{i}-\hat{j}+\hat{k}$,$\hat{i}-3\hat{j}-5\hat{k}$,and $a\hat{i}-3\hat{j}+\hat{k}$ respectively are the vertices of a right-angled triangle with $m\angle C = 90^\circ$ are:

If $4i + 7j + 8k$,$2i + 3j + 4k$ and $2i + 5j + 7k$ are the position vectors of the vertices $A$,$B$ and $C$ respectively of triangle $ABC$. The position vector of the point where the bisector of angle $A$ meets $BC$ is

If $\vec{i}+\vec{j}-\vec{k}, -\vec{i}+2\vec{j}+\vec{k}, \vec{j}+2\vec{k}, 2\vec{i}-\vec{j}+2\vec{k}$ are the position vectors of four points $A, B, C, D$ respectively,then the shortest distance between the lines $AB$ and $CD$ is

Let $\vec{a} = \hat{i} + \hat{j} + \sqrt{2}\hat{k}$,$\vec{b} = b_{1}\hat{i} + b_{2}\hat{j} + \sqrt{2}\hat{k}$,and $\vec{c} = 5\hat{i} + \hat{j} + \sqrt{2}\hat{k}$ be three vectors such that the projection vector of $\vec{b}$ on $\vec{a}$ is $\vec{a}$. If $\vec{a} + \vec{b}$ is perpendicular to $\vec{c}$,then $|\vec{b}|$ is equal to

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three unit vectors such that $|\vec{a}-\vec{b}|^{2}+|\vec{a}-\vec{c}|^{2}=8$. Then $|\vec{a}+2\vec{b}|^{2}+|\vec{a}+2\vec{c}|^{2}$ is equal to

Let $a$,$b$,and $c$ be $3$ non-zero vectors such that no $2$ of these are collinear. If vector $a + 2b$ is collinear with $c$ and $b + 3c$ is collinear with $a$,then $a + 2b + 6c$ equals: