The perpendicular distance of the point (1,2,3) from the line (x-1)/2=(y-3)/1=(z-4)/1 is ?

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Q. The perpendicular distance of the point

(2, 4, - 1)

from line

\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9}

Q. Find the perpendicular distance of a point

(1, 0, 0)

from the line

\frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 10}{8}

Statement $1$ : There exist two points on the line $\frac{x - 1}{1} = \frac{y}{- 1} = \frac{z + 2}{2}$ which are at a distance of $2$ units from point $(1, 2, - 4)$ .

Statement $2$ : Perpendicular distance of point $(1, 2, - 4)$ from the line $\frac{x - 1}{1} = \frac{y}{- 1} = \frac{z + 2}{2}$ is $1$ unit.

The distance of the point (1, 3, -7) from the plane passing through the point (1, -1, -1) having normal perpendicular to both the lines
$\frac{x - 1}{1} = \frac{y + 2}{- 2} = \frac{z - 4}{3}$ and $\frac{x - 2}{2} = \frac{y + 1}{- 1} = \frac{z + 7}{- 1}$ , is

Q. find the point on the line

\frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2}

at a distance of

3 \sqrt{2}

from the point

(1, 2, 3)

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