The distance of the line $$\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$$ from the point (1, 4, 0) along the line $$\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$$ is :

The second line is $$\frac{x}{1} = \frac{y-2}{2} = \frac{z+3}{3}$$, and its direction ratios are (1, 2, 3). A line through (1, 4, 0) with this direction can be written parametrically as $$(1+t, 4+2t, 3t)$$, where $$t$$ is a real parameter.

The first line is given by $$\frac{x-2}{2} = \frac{y-6}{3} = \frac{z-3}{4} = s$$, so any point on it can be expressed as $$(2+2s, 6+3s, 3+4s)$$, where $$s$$ is a real parameter.

$$1+t = 2+2s$$ $$4+2t = 6+3s$$ $$3t = 3+4s$$

From $$1+t = 2+2s$$ we get $$t = 1+2s$$. Substituting into $$4+2t = 6+3s$$ yields $$4+2(1+2s) = 6+3s \Rightarrow 6+4s = 6+3s \Rightarrow s = 0$$, and hence $$t = 1$$. Checking in $$3t = 3+4s$$ gives $$3(1) = 3+4(0) \Rightarrow 3 = 3$$, confirming consistency.

The intersection point is then $$(1+1, 4+2, 3) = (2, 6, 3).$$

The distance from (1, 4, 0) to (2, 6, 3) is $$d = \sqrt{(2-1)^2 + (6-4)^2 + (3-0)^2} = \sqrt{1+4+9} = \sqrt{14}.$$