The straight lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and $\frac{x - 1}{2} = \frac{y - 2}{2} = \frac{z - 3}{-2}$ are

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:

The shortest distance (in units) between the lines $\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $\vec{r}=(2\hat{i}-2\hat{j}+3\hat{k})+\lambda(\hat{i}+2\hat{j})$ is

If the two lines $\frac{x - 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect at a point,then find the value of $k$.

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

The acute angle between the lines $x=-y, z=0$ and $x=0, z=0$ is