If the lines $\frac{x-k}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x+1}{3}=\frac{y+2}{2}=\frac{z+3}{1}$ are co-planar,then the value of $k$ is $.....$

The image of the point $ (1,6,3) $ in the line $ \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} $ is

If the image of the point $P(a, 2, a)$ in the line $\frac{x}{2} = \frac{y+a}{1} = \frac{z}{1}$ is $Q$ and the image of $Q$ in the line $\frac{x-2b}{2} = \frac{y-a}{1} = \frac{z+2b}{-5}$ is $P$, then $a+b$ is equal to:

If $A(3,4,5)$, $B(4,6,3)$, $C(-1,2,4)$, and $D(1,0,5)$ are points such that the angle between the lines $DC$ and $AB$ is $\theta$, then $\cos \theta$ is equal to

The Cartesian equation of a line is $3x + 1 = 6y - 2 = -z + 1$. Find its vector equation.

The angle between the lines $\frac{x}{1} = \frac{y}{0} = \frac{z}{-1}$ and $\frac{x}{3} = \frac{y}{4} = \frac{z}{5}$ is