The sum of the coordinates of one of the points on the line $\dfrac{x - 2}{1} = \dfrac{y + 3}{-2} = \dfrac{z + 5}{2}$ which is at a distance of 3 units from the point $(2, -3, -5)$ is ...

Let any point on the given line $\dfrac{x - 2}{1} = \dfrac{y + 3}{-2} = \dfrac{z + 5}{2} = \lambda$ be $(2 + \lambda, -3 - 2\lambda, -5 + 2\lambda)$.

The distance of this point from $(2, -3, -5)$ is given as $3$ units.

Using the distance formula:

$\sqrt{(2 + \lambda - 2)^2 + (-3 - 2\lambda + 3)^2 + (-5 + 2\lambda + 5)^2} = 3$

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

$\sqrt{\lambda^2 + 4\lambda^2 + 4\lambda^2} = 3$

$\sqrt{9\lambda^2} = 3$

$3|\lambda| = 3 \Rightarrow \lambda = \pm 1$

For $\lambda = 1$, the coordinates of the point are $(2 + 1, -3 - 2(1), -5 + 2(1)) = (3, -5, -3)$.
The sum of the coordinates is $3 - 5 - 3 = -5$.

For $\lambda = -1$, the coordinates of the point are $(2 - 1, -3 - 2(-1), -5 + 2(-1)) = (1, -1, -7)$.
The sum of the coordinates is $1 - 1 - 7 = -7$.

Since $-7$ is present in the given options, it is the required sum.

Answer: $-7$