Here's a clear, educational, and well-structured solution to the problem:

1. Key Concepts and Formulas

   • Foot of the Perpendicular: If $C$ is the foot of the perpendicular from point $A$ to a plane $P$, then point $C$ lies on the plane $P$, and the vector $\vec{AC}$ is perpendicular to the plane $P$. This implies $\vec{AC}$ is parallel to the plane's normal vector.
   • Normal Vector of a Plane: For a plane given by the equation $lx + my + nz = 0$, its normal vector is $\vec{N} = (l, m, n)$.
   • Equation of a Line Perpendicular to a Plane: A line passing through a point $P_0(x_0, y_0, z_0)$ and perpendicular to a plane with normal vector $\vec{N}=(l,m,n)$ can be represented parametrically as $(x_0 + tl, y_0 + tm, z_0 + tn)$, where $t$ is a scalar parameter.
   • Distance Formula in 3D: The distance between two points $C(x_1, y_1, z_1)$ and $D(x_2, y_2, z_2)$ is given by $CD = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

2. Step-by-Step Solution

   Step 1: Determine the value of 'a' and the equation of the plane. We are given point $A(a, -2a, 3)$ and its foot of perpendicular on the plane $P: lx + my + nz = 0$ is $C(0, -a, -1)$.

   • Step 1.1: Find the vector $\vec{AC}$. The vector $\vec{AC}$ connects point $A$ to point $C$: $\vec{AC} = C - A = (0-a, -a-(-2a), -1-3) = (-a, a, -4)$ Since $\vec{AC}$ is perpendicular to the plane, it is parallel to the plane's normal vector $\vec{N}=(l, m, n)$. Thus, $\vec{N}$ is proportional to $\vec{AC}$. We can write $l = -ka$, $m = ka$, and $n = -4k$ for some non-zero scalar $k$.

   • Step 1.2: Use the fact that C lies on the plane. Point $C(0, -a, -1)$ lies on the plane $lx + my + nz = 0$. Substitute its coordinates into the plane equation: $l(0) + m(-a) + n(-1) = 0$ $-ma - n = 0 \implies n = -ma$

   • Step 1.3: Solve for 'a'. Substitute the expressions for $m$ and $n$ in terms of $k$ and $a$ from Step 1.1 into the equation from Step 1.2: $-4k = -(ka)a$ $-4k = -ka^2$ Since $k \neq 0$ (otherwise $l=m=n=0$, which does not define a plane), we can divide both sides by $-k$: $4 = a^2$ The problem states $a > 0$, so we get $a = 2$.

   • Step 1.4: Determine the coordinates of A and C, and the plane's equation. With $a=2$:

     • $A = (2, -2(2), 3) = (2, -4, 3)$
     • $C = (0, -2, -1)$ The vector $\vec{AC} = (-2, 2, -4)$. We can use a simpler normal vector for the plane by dividing $\vec{AC}$ by $-2$, so $\vec{N} = (1, -1, 2)$. Since the plane passes through the origin (as $lx+my+nz=0$) and has normal vector $(1, -1, 2)$, its equation is: $1(x-0) - 1(y-0) + 2(z-0) = 0$ $x - y + 2z = 0$ (Self-check: For $C(0, -2, -1)$, $0 - (-2) + 2(-1) = 2 - 2 = 0$. The plane equation is correct.)

   Step 2: Find the coordinates of point D. Point $D$ is the foot of the perpendicular from $B(0, 4, 5)$ to the plane $x - y + 2z = 0$.

   • Step 2.1: Write the parametric equation of the line BD. The line $BD$ passes through $B(0, 4, 5)$ and is parallel to the plane's normal vector $\vec{N}=(1, -1, 2)$. The parametric equations for the line $BD$ are: $x = 0 + 1t = t$ $y = 4 - 1t = 4-t$ $z = 5 + 2t = 5+2t$ Any point on the line $BD$ can be represented as $(t, 4-t, 5+2t)$.

   • Step 2.2: Find the intersection point D with the plane. Point $D$ lies on both the line $BD$ and the plane $x - y + 2z = 0$. Substitute the parametric coordinates of $D$ into the plane equation: $(t) - (4-t) + 2(5+2t) = 0$ $t - 4 + t + 10 + 4t = 0$ $6t + 6 = 0$ $6t = -6 \implies t = -1$

   • Step 2.3: Substitute 't' back to find D's coordinates. Substitute $t = -1$ into the parametric equations for line $BD$: $x_D = -1$ $y_D = 4 - (-1) = 5$ $z_D = 5 + 2(-1) = 3$ So, the coordinates of $D$ are $(-1, 5, 3)$.

   Step 3: Calculate the length of line segment CD. We have the coordinates of $C(0, -2, -1)$ and $D(-1, 5, 3)$. Using the 3D distance formula: $CD = \sqrt{(-1-0)^2 + (5-(-2))^2 + (3-(-1))^2}$ $CD = \sqrt{(-1)^2 + (7)^2 + (4)^2}$ $CD = \sqrt{1 + 49 + 16}$ $CD = \sqrt{66}$

3. Common Mistakes & Tips

   • Sign Errors: Be meticulous with negative signs, especially when calculating vector components and differences in the distance formula.
   • Normal Vector Direction: While the normal vector can be taken in either direction (e.g., $(l,m,n)$ or $(-l,-m,-n)$), ensure consistency in its application for the plane equation and line equations.
   • Algebraic Errors: Double-check solving linear equations for 'a' and 't' to avoid arithmetic mistakes.

4. Summary

   We first utilized the given information about point $A$ and its foot of perpendicular $C$ to determine the value of the parameter '$a$' and the specific equation of the plane. This involved finding the vector $\vec{AC}$ (which is normal to the plane) and using the fact that $C$ lies on the plane. With $a=2$, the plane equation was found to be $x-y+2z=0$. Next, we found the coordinates of point $D$, the foot of the perpendicular from point $B$ to this plane, by finding the intersection of the line passing through $B$ and perpendicular to the plane, with the plane itself. Finally, we calculated the distance between $C(0, -2, -1)$ and $D(-1, 5, 3)$ using the 3D distance formula, which resulted in $\sqrt{66}$.

5. Final Answer

The length of the line segment CD is $\sqrt{66}$, which corresponds to option (D).

The final answer is $\boxed{\sqrt{66}}$.