Let $Q(a,b,c)$ be the image of the point $P(3,2,1)$ in the line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}.$ Then the distance of $Q$ from the line $\frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2}$ is

Find the angle between the pair of lines $\frac{x+3}{3}=\frac{y-1}{5}=\frac{z+3}{4}$ and $\frac{x+1}{1}=\frac{y-4}{1}=\frac{z-5}{2}$.

Let $A$ and $B$ be two points on the line $\frac{x}{1} = \frac{y}{1} = \frac{z}{-1}$. If the distance of point $P(1, 1, 1)$ from the points $A$ and $B$ is $\sqrt{3}$,then the distance between $A$ and $B$ is:

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - c}{2} = \frac{z}{1}$ intersect, then the radius of the circle $x^2 + y^2 - 4x + 10y + c = 0$ is

Let $A$ and $B$ be two distinct points on the line $L : \frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$. Both $A$ and $B$ are at a distance $2\sqrt{17}$ from the foot of the perpendicular drawn from the point $P(1, 2, 3)$ to the line $L$. If $O$ is the origin,then $\overrightarrow{OA} \cdot \overrightarrow{OB}$ is equal to:

If $(a, b, c)$ is the image of the point $(1, 2, -3)$ in the line $\frac{x+1}{2} = \frac{y-3}{-2} = \frac{z}{-1}$,then $a+b+c$ is equal to