If the shortest distance between the lines $\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ and $\frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$ is $\frac{6}{\sqrt{5}}$,then the sum of all possible values of $\lambda$ is :

Find the values of $p$ so that the lines $\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}$ and $\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles.

$P, Q, R$ and $S$ are four points with the position vectors $3i-4j+5k, 0i+0j+4k, -4i+5j+1k$ and $-3i+4j+3k$,respectively. Then,the line $PQ$ meets the line $RS$ at the point

The square of the distance of the image of the point $A(6, 1, 5)$ in the line $\frac{x-1}{3} = \frac{y}{2} = \frac{z-2}{4}$ from the origin is:

The point of intersection of the lines represented by $\overline{r}=(\overline{i}-6 \overline{j}+2 \overline{k})+t(\overline{i}+2 \overline{j}+\overline{k})$ and $\overline{r}=(4 \overline{j}+\overline{k})+s(2 \overline{i}+\overline{j}+2 \overline{k})$ is

Statement $-1$: The point $A(1, 0, 7)$ is the mirror image of the point $B(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$.
Statement $-2$: The line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ bisects the line segment joining $A(1, 0, 7)$ and $B(1, 6, 3)$.