The shortest distance between the lines $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})$ and $\vec{r} = (2\hat{i} + 4\hat{j} + 5\hat{k}) + \mu(3\hat{i} + 4\hat{j} + 5\hat{k})$ is equal to

A line passes through the point with position vector $2\hat{i} - \hat{j} + 4\hat{k}$ and is in the direction of the vector $\hat{i} + \hat{j} - 2\hat{k}$. The equation of the line in Cartesian form is:

The direction cosines of a line equally inclined with the co-ordinate axes are

The vector equation of line passing through $(-1, 3, -2)$ and perpendicular to the lines $\frac{x+4}{1} = \frac{y}{2} = \frac{z-3}{3}$ and $\frac{x+2}{-3} = \frac{y+5}{2} = \frac{z-6}{5}$ is

The vector equation of line passing through (2, -1, 3) and perpendicular to the lines

$\frac{x-2}{3} = \frac{y-1}{1} = \frac{z+2}{2}$ and $\frac{x+3}{-4} = \frac{y-5}{-3} = \frac{z+1}{2}$ is

(Here $\lambda$ is a parameter)

If the points $P(3,\mu,6), Q(-1,-1,2)$ and $R(2,8,5)$ are collinear, then the value of '$\mu$' is

Find the root of perpendicular from origin to the line $\frac{x - 1}{3} = \frac{y - 2}{-2} = \frac{z + 1}{3}$.