The distance of the point $$(-1, 9, -16)$$ from the plane $$2x + 3y - z = 5$$ measured parallel to the line $$\frac{x+4}{3} = \frac{2-y}{4} = \frac{z-3}{12}$$ is

We need to find the distance from $$(-1, 9, -16)$$ to the plane $$2x + 3y - z = 5$$, measured parallel to the line $$\frac{x+4}{3} = \frac{2-y}{4} = \frac{z-3}{12}$$.

Next, the direction ratios of the given line are $$(3, -4, 12)$$.

We then write the parametric equations of the line passing through $$(-1, 9, -16)$$ with direction ratios $$(3, -4, 12)$$:

$$x = -1 + 3t, \quad y = 9 - 4t, \quad z = -16 + 12t$$

Substituting these into the plane equation $$2x + 3y - z = 5$$ gives:

$$2(-1 + 3t) + 3(9 - 4t) - (-16 + 12t) = 5$$

$$-2 + 6t + 27 - 12t + 16 - 12t = 5$$

$$41 - 18t = 5$$

$$18t = 36$$

$$t = 2$$

At $$t = 2$$, the intersection point is

$$ (x, y, z) = (-1 + 6, 9 - 8, -16 + 24) = (5, 1, 8)$$

Finally, the distance between $$(-1, 9, -16)$$ and $$(5, 1, 8)$$ is

$$d = \sqrt{(5+1)^2 + (1-9)^2 + (8+16)^2} = \sqrt{36 + 64 + 576} = \sqrt{676} = 26.$$

Thus, the required distance is 26.