The equation of the straight line passing through the points $(a, b, c)$ and $(a - b, b - c, c - a)$ is:

The vertices $B$ and $C$ of a triangle $ABC$ lie on the line $\frac{x}{1}=\frac{1-y}{-2}=\frac{z-2}{3}$. The coordinates of $A$ and $B$ are $(1, 6, 3)$ and $(4, 9, \alpha)$ respectively, and $C$ is at a distance of $10$ units from $B$. The area (in sq. units) of $\Delta ABC$ is:

Let the values of $\lambda$ for which the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-\lambda}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is $\frac{1}{\sqrt{6}}$ be $\lambda_1$ and $\lambda_2$. Then the radius of the circle passing through the points $(0,0), (\lambda_1, \lambda_2)$ and $(\lambda_2, \lambda_1)$ is

The lines $\bar{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3 \hat{i} - \hat{j})$ and $\bar{r} = (4 \hat{i} - \hat{k}) + \mu(2 \hat{i} + 3 \hat{k})$ are

The angle between two lines $\frac{x + 1}{2} = \frac{y + 3}{2} = \frac{z - 4}{-1}$ and $\frac{x - 4}{1} = \frac{y + 4}{2} = \frac{z + 1}{2}$ is

If the lines $\frac{x - 1}{k} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and $\frac{x - 2}{3} = \frac{y - 3}{k} = \frac{z - 1}{2}$ intersect,find the value of $k$.