The equation of the plane which contains the $$y$$-axis and passes through the point $$(1, 2, 3)$$ is:

We need the equation of a plane that contains the $$y$$-axis and passes through the point $$(1, 2, 3)$$.

The general equation of a plane is $$ax + by + cz = d$$. Since the plane contains the entire $$y$$-axis, every point $$(0, y, 0)$$ lies on it. Substituting: $$by = d$$ for all $$y$$, which requires $$b = 0$$ and $$d = 0$$.

So the equation reduces to $$ax + cz = 0$$.

Since the plane passes through $$(1, 2, 3)$$: $$a + 3c = 0$$, giving $$a = -3c$$.

The equation becomes $$-3cx + cz = 0$$, and dividing by $$c$$: $$3x - z = 0$$.

The answer is Option D: $$3x - z = 0$$.