This problem requires a solid understanding of 3D Geometry, specifically finding the equation of a plane and determining the image of a point in that plane. We will break down the solution into clear, manageable steps.

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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 passing through them 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: $A(x-x_1) + B(y-y_1) + C(z-z_1) = 0$.

• Image of a Point in a Plane: Let $P(x_P, y_P, z_P)$ be a point and $Ax+By+Cz+D=0$ be the equation of a plane. Let $Q(\alpha, \beta, \gamma)$ be the image of point $P$ in the plane. The line segment $PQ$ is perpendicular to the plane, and its midpoint lies on the plane. The coordinates of the image point $Q(\alpha, \beta, \gamma)$ can be found using the formula: $\frac{\alpha - x_P}{A} = \frac{\beta - y_P}{B} = \frac{\gamma - z_P}{C} = -2 \frac{Ax_P+By_P+Cz_P+D}{A^2+B^2+C^2}$ This formula is derived from the properties of reflection: the line joining the point and its image is perpendicular to the plane, and the midpoint of this line segment lies on the plane. The ratio $-2 \frac{Ax_P+By_P+Cz_P+D}{A^2+B^2+C^2}$ represents twice the perpendicular distance from the point $P$ to the plane, scaled by the magnitude of the normal vector, with a negative sign to account for the direction.

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2. Step-by-Step Solution

Step 1: Finding the Equation of the Plane

• What and Why: To find the image of a point in a plane, we first need the explicit algebraic equation of that plane. We are given three non-collinear points $A(1,2,0)$, $B(1,4,1)$, and $C(0,5,1)$. We will use the determinant form to find the equation of the plane.

Let $(x_1, y_1, z_1) = A(1,2,0)$. First, calculate the vectors $\vec{AB}$ and $\vec{AC}$ which lie in the plane: $\vec{AB} = (x_2-x_1, y_2-y_1, z_2-z_1) = (1-1, 4-2, 1-0) = (0, 2, 1)$ $\vec{AC} = (x_3-x_1, y_3-y_1, z_3-z_1) = (0-1, 5-2, 1-0) = (-1, 3, 1)$

Now, substitute these values into the determinant formula for the plane equation: $\left| \begin{array}{ccc} x-1 & y-2 & z-0 \\ 0 & 2 & 1 \\ -1 & 3 & 1 \end{array} \right| = 0$

Expand the determinant along the first row: $(x-1) \left| \begin{array}{cc} 2 & 1 \\ 3 & 1 \end{array} \right| - (y-2) \left| \begin{array}{cc} 0 & 1 \\ -1 & 1 \end{array} \right| + z \left| \begin{array}{cc} 0 & 2 \\ -1 & 3 \end{array} \right| = 0$ $(x-1)(2 \times 1 - 1 \times 3) - (y-2)(0 \times 1 - 1 \times (-1)) + z(0 \times 3 - 2 \times (-1)) = 0$ $(x-1)(2-3) - (y-2)(0+1) + z(0+2) = 0$ $(x-1)(-1) - (y-2)(1) + z(2) = 0$ $-x+1 -y+2 +2z = 0$ $-x -y +2z +3 = 0$ Multiplying the entire equation by -1 to obtain a standard form where the coefficient of $x$ is positive: $x+y-2z-3 = 0$ This is the equation of the plane. From this, we identify the coefficients $A=1$, $B=1$, $C=-2$, and $D=-3$.

Step 2: Finding the Image of Point P in the Plane

• What and Why: The problem asks for the coordinates of the image point $Q(\alpha, \beta, \gamma)$. We will use the standard formula for the image of a point in a plane, applying the point $P(1,2,6)$ and the plane $x+y-2z-3=0$.

We use the image formula: $\frac{\alpha - x_P}{A} = \frac{\beta - y_P}{B} = \frac{\gamma - z_P}{C} = -2 \frac{Ax_P+By_P+Cz_P+D}{A^2+B^2+C^2}$ Here, $(x_P, y_P, z_P) = (1,2,6)$ and the plane coefficients are $(A,B,C,D) = (1,1,-2,-3)$.

First, calculate the value of the common ratio on the right-hand side. Let's call this ratio $k$: $k = -2 \frac{1(1)+1(2)+(-2)(6)+(-3)}{1^2+1^2+(-2)^2}$ Substitute the coordinates of P and the plane coefficients: $k = -2 \frac{1+2-12-3}{1+1+4}$ Perform the arithmetic in the numerator and denominator: $k = -2 \frac{3-15}{6}$ $k = -2 \frac{-12}{6}$ $k = -2(-2)$ $k = 4$

Now, equate each part of the formula to $k=4$ to find $\alpha, \beta, \gamma$:

1. For $\alpha$: $\frac{\alpha - 1}{1} = 4$ $\alpha - 1 = 4$ $\alpha = 5$
2. For $\beta$: $\frac{\beta - 2}{1} = 4$ $\beta - 2 = 4$ $\beta = 6$
3. For $\gamma$: $\frac{\gamma - 6}{-2} = 4$ $\gamma - 6 = -8$ $\gamma = -2$

So, the image point $Q$ is $(\alpha, \beta, \gamma) = (5, 6, -2)$.

Step 3: Calculating $(\alpha^2+\beta^2+\gamma^2)$

• What and Why: This is the final quantity requested by the problem. We substitute the coordinates of $Q$ we just found into the expression.

Substitute the values of $\alpha=5$, $\beta=6$, and $\gamma=-2$: $\alpha^2+\beta^2+\gamma^2 = (5)^2 + (6)^2 + (-2)^2$ $= 25 + 36 + 4$ $= 61 + 4$ $= 65$

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3. Common Mistakes & Tips

• Collinearity Check: Before finding the plane equation, ensure the three given points are non-collinear. If they were collinear, they would not define a unique plane. This can be checked by verifying if the vectors $\vec{AB}$ and $\vec{AC}$ are parallel (i.e., one is a scalar multiple of the other).
• Sign Errors: Be very careful with signs, especially when expanding determinants for the plane equation and when substituting values into the image formula. A common mistake is forgetting the negative sign in the image formula's right-hand side.
• Image vs. Projection: Remember that the formula for the image of a point has a factor of $-2$, while the formula for the foot of the perpendicular (projection) has a factor of $-1$. Confusing these is a frequent error.

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4. Summary

In this problem, we systematically applied two fundamental concepts of 3D geometry. First, we derived the equation of the plane by using the determinant form with the three given points $A(1,2,0)$, $B(1,4,1)$, and $C(0,5,1)$, resulting in the plane equation $x+y-2z-3=0$. Second, we used the derived plane equation and the given point $P(1,2,6)$ to find its image $Q(\alpha, \beta, \gamma)$ using the standard formula for reflection. This yielded the coordinates $Q(5,6,-2)$. Finally, we calculated the required expression $(\alpha^2+\beta^2+\gamma^2)$ using these coordinates.

The final calculated value is $65$.

The final answer is $\boxed{\text{65}}$, which corresponds to option (D).