This problem combines concepts from 3D Geometry and Vector Algebra to determine the coordinates of an image point and then calculate the area of a triangle formed by three points. A key observation in this problem is the relationship between the given points and the plane.

1. Key Concepts and Formulas

   • Image of a point in a plane: The image $Q(x,y,z)$ of a point $P(x_1, y_1, z_1)$ in the plane $ax+by+cz+d=0$ is given by the formula: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = \frac{-2(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}$ This formula provides a direct method to find the reflection of a point across a plane. The line segment connecting the point to its image is perpendicular to the plane, and its midpoint lies on the plane.

   • Area of a triangle using vectors: If $\overrightarrow{A}$ and $\overrightarrow{B}$ are two vectors representing two adjacent sides of a triangle originating from the same vertex (e.g., $\overrightarrow{PQ}$ and $\overrightarrow{PR}$), then the area of the triangle is given by: $\text{Area of } \triangle PQR = \frac{1}{2} |\overrightarrow{PQ} \times \overrightarrow{PR}|$ where $|\overrightarrow{PQ} \times \overrightarrow{PR}|$ denotes the magnitude of the cross product of the two vectors. The cross product of two vectors $\vec{A} = A_x\hat{\mathbf{i}} + A_y\hat{\mathbf{j}} + A_z\hat{\mathbf{k}}$ and $\vec{B} = B_x\hat{\mathbf{i}} + B_y\hat{\mathbf{j}} + B_z\hat{\mathbf{k}}$ is calculated as: $\vec{A} \times \vec{B} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}$

2. Step-by-Step Solution

   Step 2.1: Finding the Coordinates of Point Q (Image of P)

   Our first task is to find the coordinates of point $Q$, which is the image of $P(1,2,3)$ in the plane $2x-y+z=9$.

   • Identify the given point P and the plane equation: The point $P(x_1, y_1, z_1)$ is $(1,2,3)$. The equation of the plane is $2x-y+z=9$, which can be written as $2x-y+z-9=0$. Comparing this with the general form $ax+by+cz+d=0$, we get the coefficients: $a=2$, $b=-1$, $c=1$, and $d=-9$.

   • Apply the image formula: Let $Q(x,y,z)$ be the image of $P(1,2,3)$. We substitute the values into the formula: $\frac{x-1}{2} = \frac{y-2}{-1} = \frac{z-3}{1} = \frac{-2(2(1) + (-1)(2) + (1)(3) + (-9))}{2^2 + (-1)^2 + 1^2}$ Explanation: This formula directly relates the coordinates of the image point to the original point and the plane's normal vector by considering the perpendicular distance from the point to the plane.

   • Calculate the numerator and denominator of the RHS: The expression inside the parenthesis in the numerator ($ax_1+by_1+cz_1+d$) is: $2(1) + (-1)(2) + (1)(3) - 9 = 2 - 2 + 3 - 9 = -6$. The denominator ($a^2+b^2+c^2$) is: $2^2 + (-1)^2 + 1^2 = 4 + 1 + 1 = 6$.

   • Simplify the Right Hand Side (RHS): $\frac{x-1}{2} = \frac{y-2}{-1} = \frac{z-3}{1} = \frac{-2(-6)}{6} = \frac{12}{6} = 2$

   • Solve for x, y, and z: Equating each part of the expression to 2:

     1. $\frac{x-1}{2} = 2 \implies x-1 = 4 \implies x = 5$
     2. $\frac{y-2}{-1} = 2 \implies y-2 = -2 \implies y = 0$
     3. $\frac{z-3}{1} = 2 \implies z-3 = 2 \implies z = 5$ Thus, the coordinates of point $Q$ are $(5,0,5)$.

   Step 2.2: Calculating the Vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$

   To find the area of $\triangle PQR$ using the cross product, we need two vectors originating from a common vertex. We'll choose vertex $P$ and calculate vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$.

   • Recall the coordinates: $P(1,2,3)$ $Q(5,0,5)$ $R(6,10,7)$

   • Calculate $\overrightarrow{PQ}$: A vector from point $A(x_A, y_A, z_A)$ to $B(x_B, y_B, z_B)$ is $\overrightarrow{AB} = (x_B-x_A)\hat{\mathbf{i}} + (y_B-y_A)\hat{\mathbf{j}} + (z_B-z_A)\hat{\mathbf{k}}$. $\overrightarrow{PQ} = (5-1)\hat{\mathbf{i}} + (0-2)\hat{\mathbf{j}} + (5-3)\hat{\mathbf{k}} = 4\hat{\mathbf{i}} - 2\hat{\mathbf{j}} + 2\hat{\mathbf{k}}$

   • Calculate $\overrightarrow{PR}$: $\overrightarrow{PR} = (6-1)\hat{\mathbf{i}} + (10-2)\hat{\mathbf{j}} + (7-3)\hat{\mathbf{k}} = 5\hat{\mathbf{i}} + 8\hat{\mathbf{j}} + 4\hat{\mathbf{k}}$ Explanation: These two vectors form two sides of the triangle $\triangle PQR$ originating from vertex $P$, which are necessary for the cross product method.

   Step 2.3: Calculating the Cross Product $\overrightarrow{PQ} \times \overrightarrow{PR}$

   Now we compute the cross product of the two vectors found in the previous step.

   • Set up the determinant: $\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 4 & -2 & 2 \\ 5 & 8 & 4 \end{vmatrix}$

   • Expand the determinant: $\overrightarrow{PQ} \times \overrightarrow{PR} = \hat{\mathbf{i}}((-2)(4) - (2)(8)) - \hat{\mathbf{j}}((4)(4) - (2)(5)) + \hat{\mathbf{k}}((4)(8) - (-2)(5))$ $\overrightarrow{PQ} \times \overrightarrow{PR} = \hat{\mathbf{i}}(-8 - 16) - \hat{\mathbf{j}}(16 - 10) + \hat{\mathbf{k}}(32 - (-10))$ $\overrightarrow{PQ} \times \overrightarrow{PR} = \hat{\mathbf{i}}(-24) - \hat{\mathbf{j}}(6) + \hat{\mathbf{k}}(32 + 10)$ $\overrightarrow{PQ} \times \overrightarrow{PR} = -24\hat{\mathbf{i}} - 6\hat{\mathbf{j}} + 42\hat{\mathbf{k}}$ Explanation: The magnitude of this resultant vector is twice the area of the triangle.

   Step 2.4: Calculating the Area of Triangle PQR and its Square

   Finally, we use the magnitude of the cross product to find the area and then square it.

   • Calculate the magnitude of the cross product: Let $\vec{V} = \overrightarrow{PQ} \times \overrightarrow{PR} = -24\hat{\mathbf{i}} - 6\hat{\mathbf{j}} + 42\hat{\mathbf{k}}$. The magnitude of $\vec{V}$ is: $|\vec{V}| = \sqrt{(-24)^2 + (-6)^2 + (42)^2}$ $|\vec{V}| = \sqrt{576 + 36 + 1764}$ $|\vec{V}| = \sqrt{2376}$

   • Calculate the area of $\triangle PQR$: $\text{Area of } \triangle PQR = \frac{1}{2} |\overrightarrow{PQ} \times \overrightarrow{PR}| = \frac{1}{2} \sqrt{2376}$ We can simplify $\sqrt{2376} = \sqrt{36 \times 66} = 6\sqrt{66}$. So, $\text{Area of } \triangle PQR = \frac{1}{2} (6\sqrt{66}) = 3\sqrt{66}$.

   • Calculate the square of the area: The question asks for the square of the area of the triangle. $(\text{Area of } \triangle PQR)^2 = (3\sqrt{66})^2 = 3^2 \times (\sqrt{66})^2 = 9 \times 66 = 594$ Self-check/Important Observation: Notice that R(6,10,7) lies on the plane $2x-y+z=9$ because $2(6) - 10 + 7 = 12 - 10 + 7 = 9$. Also, the midpoint of PQ, M(3,1,4), lies on the plane. Since $\overrightarrow{PQ}$ is perpendicular to the plane and $\overrightarrow{MR}$ lies in the plane, $\triangle PMR$ is a right-angled triangle at M. The area of $\triangle PQR$ can also be calculated as $PM \times MR$. $PM = \frac{1}{2}|\overrightarrow{PQ}| = \frac{1}{2}\sqrt{4^2+(-2)^2+2^2} = \frac{1}{2}\sqrt{24} = \sqrt{6}$. $MR = |\overrightarrow{MR}| = \sqrt{(6-3)^2+(10-1)^2+(7-4)^2} = \sqrt{3^2+9^2+3^2} = \sqrt{9+81+9} = \sqrt{99}$. Area $= \sqrt{6} \times \sqrt{99} = \sqrt{594}$. Square of Area $= 594$. Both methods yield the same result, confirming the calculation.

3. Common Mistakes & Tips

   • Sign Errors in Image Formula: Be meticulous with signs, especially the $-2$ factor and the $d$ term in the image formula.
   • Cross Product Calculation: Errors often occur during the expansion of the determinant for the cross product, particularly with the negative sign for the $\hat{\mathbf{j}}$ component. Double-check each component calculation.
   • Vector Subtraction: Ensure coordinates are subtracted in the correct order (e.g., $Q-P$ for $\overrightarrow{PQ}$).
   • Properties of Image Point: Remember that the line connecting a point to its image is perpendicular to the plane, and its midpoint lies on the plane. Any point on the plane is equidistant from the point and its image. This can sometimes simplify area calculations if a vertex lies on the plane.

4. Summary

   To find the square of the area of $\triangle PQR$, we first determined the coordinates of point $Q$ (the image of $P$) using the specific formula for the image of a point in a plane, yielding $Q(5,0,5)$. Next, we formed two vectors originating from a common vertex, $\overrightarrow{PQ} = (4,-2,2)$ and $\overrightarrow{PR} = (5,8,4)$. We then calculated their cross product, $\overrightarrow{PQ} \times \overrightarrow{PR} = -24\hat{\mathbf{i}} - 6\hat{\mathbf{j}} + 42\hat{\mathbf{k}}$. Finally, we found the magnitude of this cross product, divided it by 2 to get the area, and then squared the result. A crucial observation was that point R lies on the plane, which provides an alternative method to verify the area calculation.

The final answer is $\boxed{594}$.