Key Concepts and Formulas

1. Parametric Equation of a Line in 3D: A line passing through two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ can be represented parametrically. First, find the direction vector $\vec{d} = (x_2-x_1, y_2-y_1, z_2-z_1)$. Then, any point $R(x,y,z)$ on the line can be expressed as: $x = x_1 + \lambda (x_2-x_1)$ $y = y_1 + \lambda (y_2-y_1)$ $z = z_1 + \lambda (z_2-z_1)$ where $\lambda$ is a scalar parameter. This form is essential for representing any point on the line using a single variable.

2. Distance Formula in 3D: The distance between two points $A(x_A, y_A, z_A)$ and $B(x_B, y_B, z_B)$ is given by: $AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2 + (z_B - z_A)^2}$ It's often convenient to work with the squared distance, $AB^2$, to avoid square roots during intermediate calculations.

3. Distance of a Point from the Origin: For a point $R(x,y,z)$, its distance from the origin $O(0,0,0)$ is $OR = \sqrt{x^2 + y^2 + z^2}$. The squared distance is $OR^2 = x^2 + y^2 + z^2$.

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Step-by-Step Solution

Step 1: Determine the Parametric Equation of the Line Passing Through P and Q

We are given the points $P(1,-2,3)$ and $Q(5,-4,7)$. First, we find the direction vector of the line, $\vec{PQ}$. This vector represents the direction of the line: $\vec{PQ} = Q - P = (5-1, -4-(-2), 7-3) = (4, -2, 4)$ So, the direction ratios of the line are $(4, -2, 4)$.

Now, we can write the parametric equations for any point $R(x,y,z)$ on the line, using point $P(1,-2,3)$ as our reference point: $x = 1 + 4\lambda$ $y = -2 - 2\lambda$ $z = 3 + 4\lambda$ This means any point $R$ on the line can be uniquely identified by a specific value of $\lambda$, and its coordinates are $(1+4\lambda, -2-2\lambda, 3+4\lambda)$. This parametric form is crucial as it allows us to apply further conditions and solve for $\lambda$.

Step 2: Use the Distance Condition to Find the Value(s) of $\lambda$

The problem states that the point $(\alpha, \beta, \gamma)$, which is a point $R$ on the line, is at a distance of 9 units from point $P(1,-2,3)$. Let $R$ be $(1+4\lambda, -2-2\lambda, 3+4\lambda)$. The distance $PR$ is 9. We use the squared distance formula to simplify calculations: $PR^2 = ((1+4\lambda) - 1)^2 + ((-2-2\lambda) - (-2))^2 + ((3+4\lambda) - 3)^2$ $PR^2 = (4\lambda)^2 + (-2\lambda)^2 + (4\lambda)^2$ Since $PR=9$, $PR^2 = 9^2 = 81$. $16\lambda^2 + 4\lambda^2 + 16\lambda^2 = 81$ $36\lambda^2 = 81$ Solving for $\lambda^2$: $\lambda^2 = \frac{81}{36} = \frac{9}{4}$ Taking the square root gives two possible values for $\lambda$: $\lambda = \pm \frac{3}{2}$ We obtain two values for $\lambda$ because there are typically two points on a line that are a fixed distance from a given point on that line – one on each side of the given point along the line.

Step 3: Find the Coordinates of the Two Candidate Points

We substitute each value of $\lambda$ back into the parametric equations of the line to find the coordinates of the two candidate points.

Case 1: For $\lambda = \frac{3}{2}$ $x = 1 + 4\left(\frac{3}{2}\right) = 1 + 6 = 7$ $y = -2 - 2\left(\frac{3}{2}\right) = -2 - 3 = -5$ $z = 3 + 4\left(\frac{3}{2}\right) = 3 + 6 = 9$ Let this point be $R_1(7, -5, 9)$.

Case 2: For $\lambda = -\frac{3}{2}$ $x = 1 + 4\left(-\frac{3}{2}\right) = 1 - 6 = -5$ $y = -2 - 2\left(-\frac{3}{2}\right) = -2 + 3 = 1$ $z = 3 + 4\left(-\frac{3}{2}\right) = 3 - 6 = -3$ Let this point be $R_2(-5, 1, -3)$. These are the two points on the line that are 9 units away from $P$.

Step 4: Determine Which Point is Farther from the Origin

The problem specifies that the point $(\alpha, \beta, \gamma)$ is "farther from the origin". We calculate the squared distance of both $R_1$ and $R_2$ from the origin $O(0,0,0)$ and compare them.

For $R_1(7, -5, 9)$: Distance from origin squared, $OR_1^2$: $OR_1^2 = 7^2 + (-5)^2 + 9^2 = 49 + 25 + 81 = 155$

For $R_2(-5, 1, -3)$: Distance from origin squared, $OR_2^2$: $OR_2^2 = (-5)^2 + 1^2 + (-3)^2 = 25 + 1 + 9 = 35$

Comparing the squared distances, $155 > 35$. Therefore, $R_1(7, -5, 9)$ is farther from the origin than $R_2(-5, 1, -3)$. So, the point $(\alpha, \beta, \gamma)$ is $(7, -5, 9)$.

Step 5: Calculate $\alpha^2+\beta^2+\gamma^2$

We have determined that $(\alpha, \beta, \gamma) = (7, -5, 9)$. We need to calculate $\alpha^2+\beta^2+\gamma^2$: $\alpha^2+\beta^2+\gamma^2 = 7^2 + (-5)^2 + 9^2$ $= 49 + 25 + 81$ $= 155$ Notice that this is exactly the squared distance of point $R_1$ from the origin, $OR_1^2$, which we calculated in Step 4.

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Common Mistakes & Tips

• Forgetting $\pm$ for $\lambda$: Always remember that solving $\lambda^2 = k$ yields $\lambda = \pm \sqrt{k}$. Ignoring the negative root would lead to missing one of the candidate points.
• Working with Squared Distances: When comparing distances or when the final answer involves sums of squares, it's often more efficient and less prone to calculation errors to work with squared distances ($d^2$) rather than actual distances ($d$).
• Careful with Signs: Be meticulous with signs, especially when substituting negative coordinates or values of $\lambda$ into the equations.

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Summary

This problem required us to first establish the parametric equation of the line passing through the given points $P$ and $Q$. Then, we used the condition that the desired point is 9 units away from $P$ to find two possible values for the parameter $\lambda$, leading to two candidate points on the line. Finally, we applied the condition that the point must be farther from the origin to select the correct point $(\alpha, \beta, \gamma)$ and then calculated the required sum of squares.

The final answer is $\boxed{155}$, which corresponds to option (A).