Step 1: Understanding the Question:
The distance is not the perpendicular distance. It is the distance between point $A$ and the point where a line passing through $A$ (parallel to the given line) intersects the plane.
Step 2: Key Formula or Approach:
Equation of line through $A(x_1, y_1, z_1)$ parallel to line with d.r.s $(a, b, c)$ is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = r$.
Step 3: Detailed Explanation:
Line through $A(3, -4, 5)$ parallel to $\frac{x}{2} = \frac{y}{1} = \frac{z}{-2}$ is:
$\frac{x-3}{2} = \frac{y+4}{1} = \frac{z-5}{-2} = r$.
Any point $P$ on this line is $(2r+3, r-4, -2r+5)$.
This point lies on the plane $2x + 5y - 6z = 16$:
$2(2r+3) + 5(r-4) - 6(-2r+5) = 16$
$4r + 6 + 5r - 20 + 12r - 30 = 16$
$21r - 44 = 16 \implies 21r = 60 \implies r = 60/21 = 20/7$.
The distance $AP = \sqrt{(x-x_1)^2 + (y-y_1)^2 + (z-z_1)^2}$
$AP = \sqrt{(2r)^2 + (r)^2 + (-2r)^2} = \sqrt{4r^2 + r^2 + 4r^2} = \sqrt{9r^2} = 3|r|$.
Distance $= 3 \times \frac{20}{7} = \frac{60}{7}$ units.
Step 4: Final Answer:
The distance is $\frac{60}{7}$ units.