Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three vectors such that $|\vec{a}|=3, |\vec{b}|=4, |\vec{c}|=5$ and each one of them is perpendicular to the sum of the other two. Find $|\vec{a}+\vec{b}+\vec{c}|$.

The shortest distance between the lines $r = 3i + 5j + 7k + \lambda(i + 2j + k)$ and $r = -i - j - k + \mu(7i - 6j + k)$ is

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

If $a \cdot b = 0$,then:

If $4\hat{i}+\hat{j}-\hat{k}$ and $3\hat{i}+m\hat{j}+2\hat{k}$ are at a right angle,then $m = $

For non-zero vectors $a$ and $b$, if $|a+b| < |a-b|$, then $a$ and $b$ are