The equation of the line through the point (0, 1, 2) and perpendicular to the line $$\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{-2}$$ is:

We need the equation of the line through $$(0, 1, 2)$$ that is perpendicular to the line $$\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{-2}$$.

The given line has direction ratios $$(2, 3, -2)$$ and passes through $$(1, -1, 1)$$. A general point on this line is $$(1 + 2t,\; -1 + 3t,\; 1 - 2t)$$.

The vector from $$(0, 1, 2)$$ to this general point is $$(1 + 2t,\; -2 + 3t,\; -1 - 2t)$$.

For the line joining $$(0, 1, 2)$$ to the foot of the perpendicular to be perpendicular to the given line, the dot product with $$(2, 3, -2)$$ must be zero: $$2(1 + 2t) + 3(-2 + 3t) + (-2)(-1 - 2t) = 0$$.

Expanding: $$2 + 4t - 6 + 9t + 2 + 4t = 0$$, which gives $$17t - 2 = 0$$, so $$t = \frac{2}{17}$$.

The foot of the perpendicular on the given line is $$\left(1 + \frac{4}{17},\; -1 + \frac{6}{17},\; 1 - \frac{4}{17}\right) = \left(\frac{21}{17},\; \frac{-11}{17},\; \frac{13}{17}\right)$$.

The direction vector of the required line is from $$(0, 1, 2)$$ to the foot: $$\left(\frac{21}{17},\; \frac{-11}{17} - 1,\; \frac{13}{17} - 2\right) = \left(\frac{21}{17},\; \frac{-28}{17},\; \frac{-21}{17}\right)$$.

This is proportional to $$(21, -28, -21)$$, which simplifies to $$(3, -4, -3)$$ after dividing by 7. Writing with the opposite sign, the direction ratios are $$(-3, 4, 3)$$.

Therefore, the equation of the line through $$(0, 1, 2)$$ with direction ratios $$(-3, 4, 3)$$ is $$\frac{x}{-3} = \frac{y - 1}{4} = \frac{z - 2}{3}$$.