​$$\text{The distance between the lines } \vec{r} = \hat{i} - 2\hat{j} + 3\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) \\\text{and } \vec{r} = 3\hat{i} - 2\hat{j} + 1\hat{k} + \mu(4\hat{i} + 6\hat{j} + 12\hat{k}) \text{ is:}$$​​

The point (a, b, 0) lie on:

Find the angle between the planes $$x+y+z=1$$​ and $$x-2y+3z=1$$.

If a line makes an angle of 60°, 135°, 120° with the positive x, y, z-axis, respectively, then find the direction cosines.

The equation of the plane that passes through (1, -1 2) and has direction ratios (1, 2, 3) is:

Find the distance between two points (2, 6, 5) and (2, 3, 9).

Which of the following represents direction cosines of the line?

Find mid point of (4, 3, 6) and (6, 5, 12).

​$$\text { The angle between the lines } \frac{x-4}{2}=\frac{y}{1}=\frac{z+1}{-2}, \frac{x-1}{4}=\frac{y+1}{-4}=\frac{z-2}{2} \text { is: }$$​​

​$$\text { Find the angle between the lines } \frac{x-3}{1}=\frac{y-2}{2}=\frac{z+1}{2} \text { and } \frac{x-0}{3}=\frac{y-5}{2}=\frac{z-2}{6} \text {. }$$​​

If P (2, 3, 4), Q (5, 8, 7) and R (-1, -2, 1) are collinear, then R divides PQ in the ratio: