The distance of the point $(1,0,-3)$ from the plane $x-y-z=9$ measured parallel to the line $\dfrac{x-2}{2} = \dfrac{y+2}{3} = \dfrac{z-6}{-6}$ is

Let the given point be $P(1, 0, -3)$.

The equation of the line passing through $P$ and parallel to the given line $\dfrac{x-2}{2} = \dfrac{y+2}{3} = \dfrac{z-6}{-6}$ is

$\dfrac{x-1}{2} = \dfrac{y-0}{3} = \dfrac{z+3}{-6} = r$

Any point on this line is of the form $Q(2r+1, 3r, -6r-3)$.

Since $Q$ lies on the plane $x-y-z=9$, substituting the coordinates of $Q$ in the plane equation gives

$(2r+1) - (3r) - (-6r-3) = 9$

$2r + 1 - 3r + 6r + 3 = 9$

$5r + 4 = 9 \Rightarrow 5r = 5 \Rightarrow r = 1$

Substituting $r=1$, the coordinates of $Q$ are $(3, 3, -9)$.

The required distance is the distance between $P$ and $Q$:

$PQ = \sqrt{(3-1)^2 + (3-0)^2 + (-9 - (-3))^2}$

$PQ = \sqrt{2^2 + 3^2 + (-6)^2}$

$PQ = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ units.

Answer: $7$ units