The distance of the point $$(-1, 9, -16)$$ from the plane $$2x + 3y - z = 5$$ measure 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 $$\frac{x+4}{3} = \frac{2-y}{4} = \frac{z-3}{12}$$. Since the direction ratios of this line are $$(3,-4,12)$$, a line through $$(-1,9,-16)$$ parallel to it can be written parametrically as $$x = -1 + 3t,\quad y = 9 - 4t,\quad z = -16 + 12t$$.

Substituting these into the plane equation $$2x+3y-z=5$$ gives $$2(-1+3t)+3(9-4t)-(-16+12t)=5$$, which simplifies to $$-2+6t+27-12t+16-12t=5$$ and thus $$41 - 18t = 5$$. Solving yields $$18t = 36$$ and hence $$t = 2$$.

Next, putting $$t=2$$ back into the parametric equations gives the intersection point $$(x,y,z) = (-1+6,\,9-8,\,-16+24) = (5,1,8)$$. From the above, the distance between $$(-1,9,-16)$$ and $$(5,1,8)$$ is $$\sqrt{(5+1)^2 + (1-9)^2 + (8+16)^2} = \sqrt{36+64+576} = \sqrt{676} = 26$$.

Therefore, the correct answer is Option C: $$26$$.