1. Key Concepts and Formulas

To determine the mirror image of a point $P(x_1, y_1, z_1)$ with respect to a line $L$, we utilize two fundamental geometric principles:

• Foot of the Perpendicular (N): The line segment connecting the original point $P$ and its mirror image $P'(x_2, y_2, z_2)$ is perpendicular to the line $L$. The point where this perpendicular segment intersects the line $L$ is called the foot of the perpendicular, let's denote it as $N$.
• Midpoint Property: The foot of the perpendicular $N$ acts as the midpoint of the line segment $PP'$. This means $N$ is equidistant from $P$ and $P'$.

Key Formulas and Concepts Used:

• Parametric Form of a Line: A line $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$ can be expressed in parametric form as $(x_0+a\lambda, y_0+b\lambda, z_0+c\lambda)$, where $\lambda$ is a scalar parameter.
• Direction Vector of a Line: The vector $\vec{d} = a\hat{i} + b\hat{j} + c\hat{k}$ represents the direction of the line.
• Perpendicularity Condition: Two vectors $\vec{u}$ and $\vec{v}$ are perpendicular if and only if their dot product is zero: $\vec{u} \cdot \vec{v} = 0$.
• Midpoint Formula: The midpoint of a line segment connecting $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)$.

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2. Step-by-Step Solution

Our goal is to find $P'(\alpha, \beta, \gamma)$ for the given point $P(3,4,9)$ and line $L: \frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}$.

Step 1: Represent the Line and a General Point on It

First, we convert the symmetric equation of the line $L$ into its parametric form to represent any arbitrary point $N$ lying on it. This is crucial for setting up vector equations.

$\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1} = \lambda$ From this, the coordinates of a general point $N$ on the line $L$ in terms of $\lambda$ are:

• $x_N = 3\lambda + 1$
• $y_N = 2\lambda - 1$
• $z_N = \lambda + 2$ So, $N(3\lambda+1, 2\lambda-1, \lambda+2)$. The direction vector of the line $L$ is given by the denominators: $\vec{d} = 3\hat{i} + 2\hat{j} + 1\hat{k}$.

Step 2: Find the Foot of the Perpendicular ($N$)

The foot of the perpendicular, $N$, is the point on the line $L$ such that the vector $\vec{PN}$ is perpendicular to the line $L$.

Step 2a: Form the Vector $\vec{PN}$. We construct the vector connecting the given point $P(3,4,9)$ to the general point $N(3\lambda+1, 2\lambda-1, \lambda+2)$ on the line. $\vec{PN} = N - P = ( (3\lambda+1) - 3 )\hat{i} + ( (2\lambda-1) - 4 )\hat{j} + ( (\lambda+2) - 9 )\hat{k}$ $\vec{PN} = (3\lambda-2)\hat{i} + (2\lambda-5)\hat{j} + (\lambda-7)\hat{k}$

Step 2b: Apply the Perpendicularity Condition. Since $\vec{PN}$ is perpendicular to the line $L$, it must be perpendicular to the direction vector of the line, $\vec{d}$. Their dot product must be zero. This condition allows us to solve for the unique value of $\lambda$ that corresponds to the foot of the perpendicular. $\vec{PN} \cdot \vec{d} = 0$ Substituting the components: $(3\lambda-2)(3) + (2\lambda-5)(2) + (\lambda-7)(1) = 0$ $(9\lambda - 6) + (4\lambda - 10) + (\lambda - 7) = 0$ $9\lambda + 4\lambda + \lambda - 6 - 10 - 7 = 0$ $14\lambda - 23 = 0$ $14\lambda = 23$ $\lambda = \frac{23}{14}$

Step 2c: Calculate the Coordinates of $N$. Substitute the value of $\lambda$ back into the parametric equations for point $N$:

• $x_N = 3\left(\frac{23}{14}\right) + 1 = \frac{69}{14} + \frac{14}{14} = \frac{83}{14}$
• $y_N = 2\left(\frac{23}{14}\right) - 1 = \frac{46}{14} - \frac{14}{14} = \frac{32}{14} = \frac{16}{7}$
• $z_N = \left(\frac{23}{14}\right) + 2 = \frac{23}{14} + \frac{28}{14} = \frac{51}{14}$ So, the foot of the perpendicular is $N\left(\frac{83}{14}, \frac{16}{7}, \frac{51}{14}\right)$.

Step 3: Calculate the Mirror Image $P'(\alpha, \beta, \gamma)$

The foot of the perpendicular $N$ is the midpoint of the line segment $PP'$, where $P'(\alpha, \beta, \gamma)$ is the mirror image.

Step 3a: Apply the Midpoint Formula. Using the midpoint formula for $N$ as the midpoint of $P(3,4,9)$ and $P'(\alpha, \beta, \gamma)$:

• $N_x = \frac{P_x + P'_x}{2} \implies \frac{83}{14} = \frac{3 + \alpha}{2}$
• $N_y = \frac{P_y + P'_y}{2} \implies \frac{16}{7} = \frac{4 + \beta}{2}$
• $N_z = \frac{P_z + P'_z}{2} \implies \frac{51}{14} = \frac{9 + \gamma}{2}$

Step 3b: Solve for $\alpha, \beta, \gamma$.

• For $\alpha$: $\frac{83}{14} = \frac{3 + \alpha}{2}$ $83 = 7(3 + \alpha)$ $83 = 21 + 7\alpha$ $7\alpha = 62 \implies \alpha = \frac{62}{7}$
• For $\beta$: $\frac{16}{7} = \frac{4 + \beta}{2}$ $32 = 7(4 + \beta)$ $32 = 28 + 7\beta$ $7\beta = 4 \implies \beta = \frac{4}{7}$
• For $\gamma$: $\frac{51}{14} = \frac{9 + \gamma}{2}$ $51 = 7(9 + \gamma)$ $51 = 63 + 7\gamma$ $7\gamma = -12 \implies \gamma = -\frac{12}{7}$ So, the mirror image is $P'\left(\frac{62}{7}, \frac{4}{7}, -\frac{12}{7}\right)$.

Step 4: Final Calculation

The problem asks for the value of $14(\alpha+\beta+\gamma)$.

Step 4a: Calculate $\alpha+\beta+\gamma$. $\alpha+\beta+\gamma = \frac{62}{7} + \frac{4}{7} + \left(-\frac{12}{7}\right)$ $\alpha+\beta+\gamma = \frac{62+4-12}{7} = \frac{66-12}{7} = \frac{54}{7}$

Step 4b: Multiply by 14. $14(\alpha+\beta+\gamma) = 14 \left(\frac{54}{7}\right)$ $= 2 \times 54 = 108$

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3. Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs, especially when forming the vector $\vec{PN}$ by subtracting coordinates. For example, $(x_N - x_P)$ vs. $(x_P - x_N)$.
• Arithmetic with Fractions: Calculations involving fractions can be prone to errors. Always double-check additions, subtractions, and multiplications.
• Conceptual Clarity: Ensure a solid understanding of why the dot product is zero (perpendicularity) and why $N$ is the midpoint. These are the core principles.
• Coordinate Misreading: A common error is misreading or miscopying coordinates from the problem statement, which can lead to incorrect constant terms in the dot product equation (e.g., if $P_z=9$ was accidentally taken as $P_z=8.5$, the constant term in $14\lambda - 23 = 0$ would become $-22.5$, changing $\lambda$ to $\frac{45}{28}$ and the final answer to 102). Always verify the input values.

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4. Summary

To find the mirror image of point $P(3,4,9)$ in the line $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}$, we first expressed a general point $N$ on the line in parametric form. Then, we formed the vector $\vec{PN}$ and used the perpendicularity condition $(\vec{PN} \cdot \vec{d} = 0)$ to find the specific value of the parameter $\lambda$ that identifies the foot of the perpendicular, $N$. With $N$ determined, we leveraged the midpoint property (N is the midpoint of $P$ and its image $P'$) to calculate the coordinates $(\alpha, \beta, \gamma)$ of the mirror image. Finally, we calculated the required expression $14(\alpha+\beta+\gamma)$. Following these steps meticulously, we find the sum of coordinates $\alpha+\beta+\gamma = \frac{54}{7}$, leading to $14(\alpha+\beta+\gamma) = 108$.

5. Final Answer The final answer is $\boxed{108}$ which corresponds to option (D).