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

$\vec{a}, \vec{b}, \text{ and } \vec{c}$ are three vectors such that $|\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7$. If $\vec{a}, \vec{b}, \vec{c}$ are perpendicular to the vectors $\vec{b}+\vec{c}, \vec{c}+\vec{a}, \vec{a}+\vec{b}$ respectively,then $\sqrt{|\vec{a}+\vec{b}+\vec{c}|^2-2} = $

The cosine of the angle $A$ of the triangle with vertices $A(1, -1, 2)$,$B(6, 11, 2)$,and $C(1, 2, 6)$ 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

If the vector $a + b$ makes equal angles with vectors $a$ and $b$,then:

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}, \vec{c}=\lambda \hat{j}+\mu \hat{k}$ and $\hat{d}$ be a unit vector such that $\vec{a} \times \hat{d}=\vec{b} \times \hat{d}$ and $\vec{c} \cdot \hat{d}=1$. If $\vec{c}$ is perpendicular to $\vec{a}$,then $|3 \lambda \hat{d}+\mu \vec{c}|^2$ is equal to . . . . . . .