1. Key Concepts and Formulas

• Parametric Form of a Line: 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 as $(x_0+a\lambda, y_0+b\lambda, z_0+c\lambda)$, where $(a,b,c)$ is the direction vector of the line and $(x_0,y_0,z_0)$ is a point on the line.
• Foot of the Perpendicular: The foot of the perpendicular $Q$ from a point $P$ to a line $L$ is a point on $L$ such that the vector $\vec{PQ}$ is perpendicular to the direction vector of $L$. Mathematically, their dot product is zero: $\vec{PQ} \cdot \mathbf{d} = 0$.
• Image of a Point in a Line: If $Q$ is the foot of the perpendicular from point $P$ to line $L$, then $Q$ is the midpoint of $P$ and its image $R$. The midpoint formula states that if $Q$ is the midpoint of $P(x_1, y_1, z_1)$ and $R(x_2, y_2, z_2)$, then $Q = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)$.

2. Step-by-Step Solution

Step 1: Represent the Line Parametrically and Define the Foot of the Perpendicular

We are given the point $P = (4,4,3)$ and the line $L: \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}$. The image of $P$ in $L$ is $R = (\alpha, \beta, \gamma)$. We need to find $\alpha+\beta+\gamma$.

To find any point on the line $L$, we convert its symmetric form to a parametric form. Let each ratio be equal to a parameter $\lambda$: $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3} = \lambda$ This gives the coordinates of any point on line $L$ in terms of $\lambda$: $x = 2\lambda+1$ $y = \lambda+2$ $z = 3\lambda+1$ Let $Q$ be the foot of the perpendicular from $P$ to the line $L$. Since $Q$ lies on $L$, its coordinates can be written as: $Q = (2\lambda+1, \lambda+2, 3\lambda+1)$

Step 2: Find the Value of $\lambda$ for the Foot of the Perpendicular (Q)

The vector $\vec{PQ}$ must be perpendicular to the direction vector of the line $L$.

1. Form the vector $\vec{PQ}$: The coordinates of $P$ are $(4,4,3)$. The coordinates of $Q$ are $(2\lambda+1, \lambda+2, 3\lambda+1)$. The vector $\vec{PQ}$ is obtained by subtracting the coordinates of $P$ from $Q$: $\vec{PQ} = ((2\lambda+1) - 4)\hat{i} + ((\lambda+2) - 4)\hat{j} + ((3\lambda+1) - 3)\hat{k}$ $\vec{PQ} = (2\lambda-3)\hat{i} + (\lambda-2)\hat{j} + (3\lambda-2)\hat{k}$

2. Identify the direction vector of the line $L$: From the symmetric form of the line $L: \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}$, the direction vector is: $\mathbf{d} = 2\hat{i} + 1\hat{j} + 3\hat{k}$

3. Apply the perpendicularity condition ($\vec{PQ} \cdot \mathbf{d} = 0$): The dot product of $\vec{PQ}$ and $\mathbf{d}$ must be zero: $(2\lambda-3)(2) + (\lambda-2)(1) + (3\lambda-2)(3) = 0$ Expand and combine like terms: $(4\lambda - 6) + (\lambda - 2) + (9\lambda - 6) = 0$ $(4\lambda + \lambda + 9\lambda) + (-6 - 2 - 6) = 0$ $14\lambda - 17.5 = 0$ Solving for $\lambda$: $14\lambda = 17.5$ $\lambda = \frac{17.5}{14} = \frac{35/2}{14} = \frac{35}{28} = \frac{5}{4}$

4. Calculate the coordinates of Q: Substitute $\lambda = 5/4$ back into the parametric equations for $Q$: $Q = \left(2\left(\frac{5}{4}\right)+1, \left(\frac{5}{4}\right)+2, 3\left(\frac{5}{4}\right)+1\right)$ $Q = \left(\frac{5}{2}+1, \frac{5}{4}+\frac{8}{4}, \frac{15}{4}+\frac{4}{4}\right)$ $Q = \left(\frac{7}{2}, \frac{13}{4}, \frac{19}{4}\right)$ This is the foot of the perpendicular from $P$ to the line $L$.

Step 3: Find the Image of the Point (R)

The foot of the perpendicular $Q$ is the midpoint of the original point $P(4,4,3)$ and its image $R(\alpha, \beta, \gamma)$. Using the midpoint formula: $x_Q = \frac{x_P + x_R}{2} \quad \implies \quad \alpha = 2x_Q - x_P$ $y_Q = \frac{y_P + y_R}{2} \quad \implies \quad \beta = 2y_Q - y_P$ $z_Q = \frac{z_P + z_R}{2} \quad \implies \quad \gamma = 2z_Q - z_P$ Substitute the coordinates of $P=(4,4,3)$ and $Q=(7/2, 13/4, 19/4)$:

1. For the x-coordinate $\alpha$: $\alpha = 2\left(\frac{7}{2}\right) - 4 = 7 - 4 = 3$
2. For the y-coordinate $\beta$: $\beta = 2\left(\frac{13}{4}\right) - 4 = \frac{13}{2} - 4 = \frac{13}{2} - \frac{8}{2} = \frac{5}{2}$
3. For the z-coordinate $\gamma$: $\gamma = 2\left(\frac{19}{4}\right) - 3 = \frac{19}{2} - 3 = \frac{19}{2} - \frac{6}{2} = \frac{13}{2}$ Thus, the image of point $P(4,4,3)$ in the line $L$ is $R = \left(3, \frac{5}{2}, \frac{13}{2}\right)$.

Step 4: Calculate the Required Sum

The problem asks for the sum $\alpha+\beta+\gamma$: $\alpha+\beta+\gamma = 3 + \frac{5}{2} + \frac{13}{2}$ $\alpha+\beta+\gamma = 3 + \frac{18}{2}$ $\alpha+\beta+\gamma = 3 + 9$ $\alpha+\beta+\gamma = 12$

3. Common Mistakes & Tips

• Arithmetic Errors: Be extremely careful with calculations, especially when expanding dot products and combining terms. A small mistake in addition or subtraction can lead to an incorrect value of $\lambda$.
• Vector Direction: Ensure the direction vector of the line is correctly identified from its symmetric form. It's the denominators of the $(x-x_0)$, $(y-y_0)$, and $(z-z_0)$ terms.
• Midpoint Formula Application: Remember that the foot of the perpendicular is the midpoint, not the image itself. Double-check the application of the midpoint formula to correctly find the image coordinates.

4. Summary

To find the image of a point in a line, we first express the line in parametric form to represent the foot of the perpendicular $Q$. We then use the condition that the vector from the original point to $Q$ is perpendicular to the line's direction vector to find the specific coordinates of $Q$. Finally, $Q$ serves as the midpoint between the original point and its image, allowing us to calculate the image's coordinates using the midpoint formula. In this problem, we found the foot of the perpendicular $Q=\left(\frac{7}{2}, \frac{13}{4}, \frac{19}{4}\right)$ and the image $R=\left(3, \frac{5}{2}, \frac{13}{2}\right)$, leading to a sum of coordinates $\alpha+\beta+\gamma = 12$.

The final answer is $\boxed{\text{12}}$, which corresponds to option (A).