This problem requires a systematic approach involving three main steps in 3D Geometry: first, determining the equation of the plane from three given points; second, finding the coordinates of the foot of the perpendicular from a given point to this plane; and finally, calculating the distance of this foot from the origin.

1. Key Concepts and Formulas

• Equation of a Plane through Three Non-Collinear Points: Given three non-collinear points $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $C(x_3, y_3, z_3)$, the equation of the plane can be found using the determinant form: $\left|\begin{array}{ccc} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{array}\right| = 0$ Alternatively, one can find two vectors lying in the plane (e.g., $\vec{AB}$ and $\vec{AC}$), compute their cross product to get the normal vector $\vec{n} = (a,b,c)$, and then use the point-normal form $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$.

• Foot of the Perpendicular from a Point to a Plane: To find the foot of the perpendicular $Q(\alpha, \beta, \gamma)$ from a point $P(x_0, y_0, z_0)$ to a plane $Ax+By+Cz+D=0$:

  1. The line $PQ$ is perpendicular to the plane, so its direction ratios are $(A,B,C)$, which are the components of the normal vector.
  2. Write the parametric equation of the line $PQ$: $\frac{x-x_0}{A} = \frac{y-y_0}{B} = \frac{z-z_0}{C} = \lambda$.
  3. Express any point on the line $PQ$ in terms of $\lambda$: $(A\lambda+x_0, B\lambda+y_0, C\lambda+z_0)$.
  4. Since $Q$ lies on the plane, substitute its parametric coordinates into the plane equation and solve for $\lambda$.
  5. Substitute the value of $\lambda$ back into the parametric coordinates to find $Q$.

• Distance Formula in 3D: The distance between a point $Q(\alpha, \beta, \gamma)$ and the origin $O(0,0,0)$ is given by $OQ = \sqrt{\alpha^2 + \beta^2 + \gamma^2}$.

2. Step-by-Step Solution

Step 1: Determine the Equation of the Plane

• What we're doing: Finding the algebraic equation that defines the plane.
• Why we're doing it: The foot of the perpendicular is defined relative to this plane, so its equation is fundamental to the problem.

The plane passes through the points $A(-1, -2, -3)$, $B(9, 3, 4)$, and $C(9, -2, 1)$. First, we find two vectors lying in the plane: $\vec{AB} = (9 - (-1), 3 - (-2), 4 - (-3)) = (10, 5, 7)$ $\vec{AC} = (9 - (-1), -2 - (-2), 1 - (-3)) = (10, 0, 4)$

Now, we find the normal vector to the plane by taking the cross product of $\vec{AB}$ and $\vec{AC}$: $\vec{n} = \vec{AB} \times \vec{AC} = \left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 10 & 5 & 7 \\ 10 & 0 & 4 \end{array}\right|$ $\vec{n} = \mathbf{i}(5 \cdot 4 - 7 \cdot 0) - \mathbf{j}(10 \cdot 4 - 7 \cdot 10) + \mathbf{k}(10 \cdot 0 - 5 \cdot 10)$ $\vec{n} = \mathbf{i}(20 - 0) - \mathbf{j}(40 - 70) + \mathbf{k}(0 - 50)$ $\vec{n} = 20\mathbf{i} + 30\mathbf{j} - 50\mathbf{k}$ The normal vector is $(20, 30, -50)$, which can be simplified by dividing by 10 to $(2, 3, -5)$.

Using the point-normal form $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$ with point $A(-1, -2, -3)$ and normal vector $(2, 3, -5)$: $2(x - (-1)) + 3(y - (-2)) - 5(z - (-3)) = 0$ $2(x+1) + 3(y+2) - 5(z+3) = 0$ $2x + 2 + 3y + 6 - 5z - 15 = 0$ $2x + 3y - 5z - 7 = 0$

This is the equation of the plane.

Step 2: Find the Coordinates of the Foot of the Perpendicular $Q(\alpha, \beta, \gamma)$

• What we're doing: Determining the exact coordinates of point $Q$.
• Why we're doing it: Point $Q$ is the specific point whose distance from the origin we need to find.

We need to find the foot of the perpendicular $Q$ from point $P(3, -2, -9)$ to the plane $2x + 3y - 5z - 7 = 0$. The normal vector to the plane is $\vec{n} = (2, 3, -5)$. The line $PQ$ is perpendicular to the plane, so its direction ratios are $(2, 3, -5)$.

The parametric equation of the line $PQ$ passing through $P(3, -2, -9)$ is: $\frac{x-3}{2} = \frac{y-(-2)}{3} = \frac{z-(-9)}{-5} = \lambda$ $\frac{x-3}{2} = \frac{y+2}{3} = \frac{z+9}{-5} = \lambda$

Any point $Q$ on this line can be expressed in terms of $\lambda$: $Q(\alpha, \beta, \gamma) = (2\lambda+3, 3\lambda-2, -5\lambda-9)$.

Since $Q$ lies on the plane, its coordinates must satisfy the plane's equation: $2(2\lambda+3) + 3(3\lambda-2) - 5(-5\lambda-9) - 7 = 0$

Now, solve for $\lambda$: $4\lambda + 6 + 9\lambda - 6 + 25\lambda + 45 - 7 = 0$ Combine $\lambda$ terms and constant terms: $(4+9+25)\lambda + (6-6+45-7) = 0$ $38\lambda + 38 = 0$ $38\lambda = -38$ $\lambda = -1$

Substitute $\lambda = -1$ back into the coordinates of $Q$: $\alpha = 2(-1) + 3 = -2 + 3 = 1$ $\beta = 3(-1) - 2 = -3 - 2 = -5$ $\gamma = -5(-1) - 9 = 5 - 9 = -4$

So, the foot of the perpendicular is $Q(1, -5, -4)$.

Step 3: Calculate the Distance of $Q$ from the Origin

• What we're doing: Finding the distance from $Q$ to the origin.
• Why we're doing it: This is the final requirement of the problem statement.

We have found $Q(1, -5, -4)$. We need its distance from the origin $O(0,0,0)$. Using the 3D distance formula: $OQ = \sqrt{(1-0)^2 + (-5-0)^2 + (-4-0)^2}$ $OQ = \sqrt{1^2 + (-5)^2 + (-4)^2}$ $OQ = \sqrt{1 + 25 + 16}$ $OQ = \sqrt{42}$

3. Common Mistakes & Tips

• Sign Errors: Be extremely careful with negative signs when subtracting coordinates, expanding determinants, or substituting values. A single sign error can propagate through the entire problem.
• Arithmetic Precision: Double-check all arithmetic, especially when combining terms or solving for $\lambda$. Simplifying the plane equation by dividing by a common factor can help reduce large numbers and potential errors.
• Verification: After finding $Q$, you can quickly verify if it lies on the plane by substituting its coordinates into the plane equation. For $Q(1,-5,-4)$: $2(1) + 3(-5) - 5(-4) - 7 = 2 - 15 + 20 - 7 = 0$. This confirms $Q$ is on the plane.

4. Summary

We first found the equation of the plane passing through the three given points to be $2x + 3y - 5z - 7 = 0$. Next, we determined the coordinates of the foot of the perpendicular $Q$ from point $P(3, -2, -9)$ to this plane by setting up the parametric equation of the line $PQ$ and finding its intersection with the plane, which gave $Q(1, -5, -4)$. Finally, we calculated the distance of $Q$ from the origin, which is $\sqrt{42}$.

Based on the provided options and the derived answer, there seems to be a discrepancy. The mathematically derived answer from the given problem statement is $\sqrt{42}$, which corresponds to option (C). However, the stated Correct Answer is (A) $\sqrt{38}$. Since the instructions require the derivation to arrive at the provided correct answer, and a rigorous derivation leads to $\sqrt{42}$, there might be a typo in the question's parameters or the given correct answer. Assuming the problem intends to lead to option (A), this solution presents the mathematically correct steps for the given problem.

The final answer is $\boxed{\sqrt{42}}$ which corresponds to option (C).