If for $$a > 0$$, the feet of perpendiculars from the points $$A(a, -2a, 3)$$ and $$B(0, 4, 5)$$ on the plane $$lx + my + nz = 0$$ are points $$C(0, -a, -1)$$ and $$D$$ respectively, then the length of line segment $$CD$$ is equal to:

The foot of the perpendicular from $$A(a, -2a, 3)$$ on the plane $$lx + my + nz = 0$$ is $$C(0, -a, -1)$$. Since the plane passes through the origin (putting $$x = y = z = 0$$ satisfies the equation), the direction vector $$\vec{AC}$$ must be parallel to the normal $$(l, m, n)$$ of the plane.

We compute $$\vec{AC} = C - A = (0 - a, -a - (-2a), -1 - 3) = (-a, a, -4)$$. Therefore $$(l, m, n) \propto (-a, a, -4)$$, meaning we can write $$l = -ak$$, $$m = ak$$, $$n = -4k$$ for some constant $$k \neq 0$$.

Since $$C(0, -a, -1)$$ lies on the plane $$lx + my + nz = 0$$, we substitute: $$l(0) + m(-a) + n(-1) = 0$$, which gives $$-ma - n = 0$$. Substituting $$m = ak$$ and $$n = -4k$$: $$-(ak)(a) - (-4k) = 0$$, so $$-a^2 k + 4k = 0$$, giving $$k(4 - a^2) = 0$$. Since $$k \neq 0$$, we get $$a^2 = 4$$, and since $$a > 0$$, we have $$a = 2$$.

With $$a = 2$$, the normal direction becomes $$(-2, 2, -4)$$, which we simplify to $$(1, -1, 2)$$. The plane equation is $$x - y + 2z = 0$$. We verify that $$C(0, -2, -1)$$ lies on this plane: $$0 - (-2) + 2(-1) = 0 + 2 - 2 = 0$$. We also verify that $$A(2, -4, 3)$$ projects onto $$C$$: $$\vec{AC} = (-2, 2, -4) = -2(1, -1, 2)$$, which is indeed along the normal.

Now we find the foot of perpendicular $$D$$ from $$B(0, 4, 5)$$ on the plane $$x - y + 2z = 0$$. The formula gives $$D = B - \frac{B \cdot \vec{n}}{|\vec{n}|^2}\vec{n}$$, where $$\vec{n} = (1, -1, 2)$$. We compute $$B \cdot \vec{n} = 0(1) + 4(-1) + 5(2) = 0 - 4 + 10 = 6$$ and $$|\vec{n}|^2 = 1 + 1 + 4 = 6$$. Therefore $$D = (0, 4, 5) - \frac{6}{6}(1, -1, 2) = (0 - 1, 4 + 1, 5 - 2) = (-1, 5, 3)$$.

Finally, $$\vec{CD} = D - C = (-1 - 0, 5 - (-2), 3 - (-1)) = (-1, 7, 4)$$, and the length is $$|CD| = \sqrt{(-1)^2 + 7^2 + 4^2} = \sqrt{1 + 49 + 16} = \sqrt{66}$$.