The Cartesian equation of the line which is parallel to the line $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$ and passing through the point $(1, -3, 5)$ is:

The distance of the point $(1, 2, -4)$ from the line $\frac{x-3}{2} = \frac{y-3}{3} = \frac{z+5}{6}$ is

One vertex of a rectangular parallelepiped is at the origin $O$ and the lengths of its edges along $x, y$ and $z$ axes are $3, 4$ and $5$ units respectively. Let $P$ be the vertex $(3, 4, 5)$. Then the shortest distance between the diagonal $OP$ and an edge parallel to the $z$-axis,not passing through $O$ or $P$,is:

If the line passing through the points $(a, 1, 6)$ and $(3, 4, b)$ crosses the $yz$-plane at the point $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$,then:

Let $P(\alpha, \beta, \gamma)$ be the point on the line $\frac{x-1}{2} = \frac{y+1}{-3} = \frac{z}{1}$ at a distance $4\sqrt{14}$ from the point $(1, -1, 0)$ and nearer to the origin. Then the shortest distance between the lines $\frac{x-\alpha}{1} = \frac{y-\beta}{2} = \frac{z-\gamma}{3}$ and $\frac{x+5}{2} = \frac{y-10}{1} = \frac{z-3}{1}$ is equal to

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 $B(1, 6, 3)$.