1. Key Concepts and Formulas

• Centroid of a Triangle: For a triangle with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$, the centroid G is given by $G = \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3} \right)$.
• Parametric Form of a Line: A line defined by the intersection of two planes can be represented parametrically by expressing its coordinates in terms of a single variable (e.g., $t$).
• Distance Formula in 3D: The squared distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $d^2 = (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$. Minimizing the squared distance is equivalent to minimizing the distance itself.
• Minimization of a Quadratic Function: A quadratic function $f(t) = at^2 + bt + c$ with $a>0$ has a minimum value at $t = -b/(2a)$.

2. Step-by-Step Solution

Step 1: Determine the coordinates of the centroid G of $\Delta ABC$.

• Why: The centroid G is one of the points whose distance needs to be minimized. Its coordinates are the starting point for the distance calculation.
• Math: Given vertices $A(0, \alpha, \alpha)$, $B(\alpha, 0, \alpha)$, and $C(\alpha, \alpha, 0)$. Using the centroid formula: $G = \left( \frac{0+\alpha+\alpha}{3}, \frac{\alpha+0+\alpha}{3}, \frac{\alpha+\alpha+0}{3} \right)$ $G = \left( \frac{2\alpha}{3}, \frac{2\alpha}{3}, \frac{2\alpha}{3} \right)$

Step 2: Express the coordinates of point D parametrically.

• Why: Point D moves on a line. To calculate the distance GD, we need a general representation for any point D on this line.
• Math: The line is given by the equations $x+z-3=0$ and $y=0$. From $y=0$, we know the y-coordinate of D is $0$. From $x+z-3=0$, we have $x+z=3$. Let's introduce a parameter $t$ for $x$. So, let $x_D = t$. Then, $t+z_D=3 \implies z_D = 3-t$. Thus, any point D on the line can be represented as $D(t, 0, 3-t)$.

Step 3: Formulate the squared distance $GD^2$ as a function of the parameter $t$.

• Why: Calculating the squared distance simplifies the algebra by avoiding square roots, especially when finding the minimum. Minimizing $GD^2$ is equivalent to minimizing GD.
• Math: Using the distance formula for $G\left(\frac{2\alpha}{3}, \frac{2\alpha}{3}, \frac{2\alpha}{3}\right)$ and $D(t, 0, 3-t)$: $GD^2 = \left(t - \frac{2\alpha}{3}\right)^2 + \left(0 - \frac{2\alpha}{3}\right)^2 + \left((3-t) - \frac{2\alpha}{3}\right)^2$ $GD^2 = \left(t - \frac{2\alpha}{3}\right)^2 + \left(\frac{2\alpha}{3}\right)^2 + \left(\left(3 - \frac{2\alpha}{3}\right) - t\right)^2$ Let $K = \frac{2\alpha}{3}$ and $M = 3 - \frac{2\alpha}{3}$. The expression becomes: $GD^2 = (t-K)^2 + K^2 + (M-t)^2$ Expanding and collecting terms: $GD^2 = (t^2 - 2Kt + K^2) + K^2 + (M^2 - 2Mt + t^2)$ $GD^2 = 2t^2 - 2(K+M)t + (2K^2 + M^2)$ Substitute $K+M = \frac{2\alpha}{3} + (3 - \frac{2\alpha}{3}) = 3$. $GD^2 = 2t^2 - 2(3)t + \left(2\left(\frac{2\alpha}{3}\right)^2 + \left(3 - \frac{2\alpha}{3}\right)^2\right)$ $GD^2 = 2t^2 - 6t + \left(\frac{8\alpha^2}{9} + \left(\frac{9-2\alpha}{3}\right)^2\right)$ $GD^2 = 2t^2 - 6t + \left(\frac{8\alpha^2}{9} + \frac{81 - 36\alpha + 4\alpha^2}{9}\right)$ $GD^2 = 2t^2 - 6t + \frac{12\alpha^2 - 36\alpha + 81}{9}$ $GD^2 = 2t^2 - 6t + \frac{4\alpha^2 - 12\alpha + 27}{3}$ Let $f(t) = GD^2$.

Step 4: Find the minimum value of $GD^2$.

• Why: $f(t)$ is a quadratic function of $t$ in the form $at^2+bt+c$ with $a=2 > 0$. Its minimum value occurs at the vertex.
• Math: The minimum occurs at $t_{min} = -\frac{b}{2a} = -\frac{-6}{2(2)} = \frac{6}{4} = \frac{3}{2}$. Substitute $t_{min} = \frac{3}{2}$ back into the expression for $GD^2$: $GD_{min}^2 = 2\left(\frac{3}{2}\right)^2 - 6\left(\frac{3}{2}\right) + \frac{4\alpha^2 - 12\alpha + 27}{3}$ $GD_{min}^2 = 2\left(\frac{9}{4}\right) - 9 + \frac{4\alpha^2 - 12\alpha + 27}{3}$ $GD_{min}^2 = \frac{9}{2} - 9 + \frac{4\alpha^2 - 12\alpha + 27}{3}$ $GD_{min}^2 = -\frac{9}{2} + \frac{4\alpha^2 - 12\alpha + 27}{3}$ To combine, find a common denominator (6): $GD_{min}^2 = \frac{-9 \cdot 3}{6} + \frac{2(4\alpha^2 - 12\alpha + 27)}{6}$ $GD_{min}^2 = \frac{-27 + 8\alpha^2 - 24\alpha + 54}{6}$ $GD_{min}^2 = \frac{8\alpha^2 - 24\alpha + 27}{6}$

Step 5: Equate the minimum squared distance to the given value and solve for $\alpha$.

• Why: The problem states that the minimum length of GD is $\sqrt{\frac{57}{2}}$. We can use this to set up an equation and solve for $\alpha$.
• Math: Given $GD_{min} = \sqrt{\frac{57}{2}}$, so $GD_{min}^2 = \frac{57}{2}$. Equating our derived expression with the given value: $\frac{8\alpha^2 - 24\alpha + 27}{6} = \frac{57}{2}$ Multiply both sides by 6: $8\alpha^2 - 24\alpha + 27 = \frac{57}{2} \cdot 6$ $8\alpha^2 - 24\alpha + 27 = 57 \cdot 3$ $8\alpha^2 - 24\alpha + 27 = 171$ Rearrange into a standard quadratic equation: $8\alpha^2 - 24\alpha - 144 = 0$ Divide by 8 to simplify: $\alpha^2 - 3\alpha - 18 = 0$ Factor the quadratic equation: $(\alpha - 6)(\alpha + 3) = 0$ This gives two possible values for $\alpha$: $\alpha = 6$ or $\alpha = -3$. The problem statement specifies that $\alpha > 0$. Therefore, $\alpha = 6$.

3. Common Mistakes & Tips

• Algebraic Errors: Be meticulous with expansion and simplification of terms, especially when dealing with fractions and multiple variables.
• Forgetting Constraints: Always check the final solutions against any given constraints (e.g., $\alpha > 0$).
• Minimizing Squared Distance: Remember that minimizing $d^2$ is equivalent to minimizing $d$, and it often simplifies calculations significantly.
• Parametric Representation: When a line is given by the intersection of two planes, choose one coordinate as a parameter (e.g., $x=t$) and express the others in terms of it.

4. Summary

The solution involved a systematic approach to a 3D geometry problem. First, we calculated the centroid of the triangle using the given vertices. Next, we parameterized the line on which point D moves to represent any point D in terms of a single variable. We then formulated the squared distance between G and D as a quadratic function of this parameter. By finding the minimum of this quadratic, we obtained an expression for the minimum squared distance in terms of $\alpha$. Finally, equating this expression to the given minimum squared distance allowed us to solve for $\alpha$, yielding $\alpha=6$ as the valid solution under the constraint $\alpha > 0$.

5. Final Answer The final answer is $\boxed{6}$.