If the lines $$\frac{x-1}{1} = \frac{y-2}{2} = \frac{z+3}{1}$$ and $$\frac{x-a}{2} = \frac{y+2}{3} = \frac{z-3}{1}$$ intersects at the point $$P$$, then the distance of the point $$P$$ from the plane $$z = a$$ is:

Find the intersection of lines $$\frac{x-1}{1} = \frac{y-2}{2} = \frac{z+3}{1}$$ and $$\frac{x-a}{2} = \frac{y+2}{3} = \frac{z-3}{1}$$. To this end, we introduce parameters $$t$$ and $$s$$ and write the first line as $$(x,y,z)=(1+t,2+2t,-3+t)$$ and the second line as $$(x,y,z)=(a+2s,-2+3s,3+s)$$.

Equating corresponding coordinates yields the system $$1+t = a + 2s,\quad 2+2t = -2+3s,\quad -3+t = 3+s.$$ From the third equation we get $$t = 6 + s$$. Substituting into the second equation gives $$2 + 2(6+s) = -2 + 3s\;\Longrightarrow\;14 + 2s = -2 + 3s\;\Longrightarrow\;s = 16,$$ and hence $$t = 22$$.

Substituting $$t = 22$$ into $$(1+t,2+2t,-3+t)$$ yields the intersection point $$P = (23,46,19).$$ Then the first equation $$1+t = a + 2s$$ becomes $$23 = a + 32\;\Longrightarrow\;a = -9.$$

To find the distance from $$P$$ to the plane $$z = a = -9$$, we compute $$|z_P - a| = |19 - (-9)| = 28.$$ Therefore, the correct answer is Option 2: $$28$$.