Key Concepts and Formulas

1. Parametric Form of a Line in 3D: A line given in symmetric form $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$ can be expressed parametrically by setting each ratio equal to a parameter, say $\lambda$. Any point on the line can then be represented as $(x_0 + a\lambda, y_0 + b\lambda, z_0 + c\lambda)$. This form is essential for representing a general point on the line.

2. Distance Formula in 3D: The distance $d$ between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$ For calculations, it's often more convenient to work with the squared distance, $d^2$.

3. Centroid of a Triangle in 3D: The centroid $(\alpha, \beta, \gamma)$ of a triangle with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ is calculated as the average of their respective coordinates: $(\alpha, \beta, \gamma) = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$

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Step-by-Step Solution

Step 1: Parameterize the Given Line The equation of the line is $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$. To represent any point on this line, we set each ratio equal to a parameter $\lambda$. This converts the symmetric form into a more usable parametric form. $\frac{x+3}{8} = \lambda \implies x = 8\lambda - 3$ $\frac{y-4}{2} = \lambda \implies y = 2\lambda + 4$ $\frac{z+1}{2} = \lambda \implies z = 2\lambda - 1$ Thus, any point on the line can be represented as $A(8\lambda - 3, 2\lambda + 4, 2\lambda - 1)$. Reasoning: By expressing any point on the line in terms of a single variable $\lambda$, we can use this general point to apply the distance condition and solve for the specific values of $\lambda$ that correspond to points P and Q.

Step 2: Formulate the Distance Equation The problem states that points P and Q are on this line and are at a distance of 6 units from $R(1,2,3)$. Let $A$ be one such point (P or Q). We use the 3D distance formula, specifically the squared distance, to set up an equation. The square of the distance between $A(8\lambda - 3, 2\lambda + 4, 2\lambda - 1)$ and $R(1,2,3)$ is $6^2 = 36$. $( (8\lambda - 3) - 1 )^2 + ( (2\lambda + 4) - 2 )^2 + ( (2\lambda - 1) - 3 )^2 = 36$ Simplify the terms inside the parentheses: $(8\lambda - 4)^2 + (2\lambda + 2)^2 + (2\lambda - 4)^2 = 36$ Reasoning: This step translates the problem's geometric condition (distance of 6 units) into an algebraic equation involving $\lambda$. Squaring the distance eliminates the square root, simplifying the subsequent calculations.

Step 3: Solve the Quadratic Equation for $\lambda$ Now, we expand and simplify the equation obtained in Step 2: $(64\lambda^2 - 64\lambda + 16) + (4\lambda^2 + 8\lambda + 4) + (4\lambda^2 - 16\lambda + 16) = 36$ Combine like terms (coefficients of $\lambda^2$, $\lambda$, and constant terms): $(64+4+4)\lambda^2 + (-64+8-16)\lambda + (16+4+16) = 36$ $72\lambda^2 - 72\lambda + 36 = 36$ Subtract 36 from both sides to simplify: $72\lambda^2 - 72\lambda = 0$ Factor out $72\lambda$: $72\lambda(\lambda - 1) = 0$ This yields two possible values for $\lambda$: $\lambda = 0 \quad \text{or} \quad \lambda = 1$ Reasoning: Solving for $\lambda$ provides the specific parameter values that correspond to the points P and Q. Since the distance equation is quadratic, we expect two solutions, which aligns with the problem stating there are two such points.

Step 4: Determine the Coordinates of Points P and Q Substitute the values of $\lambda$ found in Step 3 back into the parametric equations of the line $(8\lambda - 3, 2\lambda + 4, 2\lambda - 1)$.

For $\lambda = 0$: $x_P = 8(0) - 3 = -3$ $y_P = 2(0) + 4 = 4$ $z_P = 2(0) - 1 = -1$ So, point $P = (-3, 4, -1)$.

For $\lambda = 1$: $x_Q = 8(1) - 3 = 5$ $y_Q = 2(1) + 4 = 6$ $z_Q = 2(1) - 1 = 1$ So, point $Q = (5, 6, 1)$. (The assignment of P and Q is arbitrary; swapping them will not affect the centroid calculation.) Reasoning: We now have the explicit coordinates of the two points on the line that satisfy the given distance condition. These points, along with R, form the vertices of the triangle PQR.

Step 5: Calculate the Centroid of Triangle PQR The vertices of the triangle are $P(-3, 4, -1)$, $Q(5, 6, 1)$, and $R(1, 2, 3)$. Let the centroid be $(\alpha, \beta, \gamma)$. We apply the 3D centroid formula: $\alpha = \frac{x_P + x_Q + x_R}{3} = \frac{-3 + 5 + 1}{3} = \frac{3}{3} = 1$ $\beta = \frac{y_P + y_Q + y_R}{3} = \frac{4 + 6 + 2}{3} = \frac{12}{3} = 4$ $\gamma = \frac{z_P + z_Q + z_R}{3} = \frac{-1 + 1 + 3}{3} = \frac{3}{3} = 1$ Thus, the centroid of $\triangle PQR$ is $(\alpha, \beta, \gamma) = (1, 4, 1)$. Reasoning: This is a direct application of the centroid formula, using the coordinates of all three vertices (P, Q, and the given R).

Step 6: Calculate $\alpha^2 + \beta^2 + \gamma^2$ Finally, we need to compute the value of the expression requested in the problem statement: $\alpha^2 + \beta^2 + \gamma^2 = (1)^2 + (4)^2 + (1)^2$ $= 1 + 16 + 1$ $= 18$ Reasoning: This is the final arithmetic step to arrive at the solution requested by the problem.

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Common Mistakes & Tips

• Algebraic Precision: Be meticulous when expanding squared binomials like $(8\lambda - 4)^2$. A common error is forgetting the middle term ($2ab$).
• Sign Errors: Pay close attention to negative signs, especially when substituting coordinates or simplifying expressions. A single sign error can lead to an incorrect final answer.
• Including All Vertices for Centroid: Ensure you use all three vertices (P, Q, and R) when calculating the centroid. Forgetting any vertex will lead to an incorrect result.
• Understanding Parametric Form: A strong understanding of how to convert a line's equation into parametric form is fundamental, as it allows for algebraic manipulation of points on the line.

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Summary

This problem required a sequential application of several core concepts in 3D geometry. We began by converting the line's symmetric equation into its parametric form to represent any point on it. Next, we used the distance formula to set up a quadratic equation based on the condition that points P and Q are 6 units away from R. Solving this quadratic yielded two values for the parameter $\lambda$, which in turn gave us the exact coordinates of points P and Q. Finally, with the coordinates of all three vertices (P, Q, and R), we applied the centroid formula to find $(\alpha, \beta, \gamma)$ and then calculated the required $\alpha^2 + \beta^2 + \gamma^2$. The problem highlights the interconnectedness of these concepts and the importance of systematic algebraic manipulation.

The final answer is $\boxed{18}$. which corresponds to option (A).