1. Key Concepts and Formulas

• Image of a Point in a Line: If $P'$ is the image of a point $P$ with respect to a line $L$, and $A$ is the foot of the perpendicular from $P$ to $L$, then $A$ is the midpoint of the line segment $PP'$. Also, the line segment $PA$ is perpendicular to the line $L$.
• Parametric Form of a Line: A line given in symmetric form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$ can be written in parametric form as $x=x_1+a\lambda$, $y=y_1+b\lambda$, $z=z_1+c\lambda$, where $(x_1,y_1,z_1)$ is a point on the line and $\vec{d}=(a,b,c)$ is its direction vector.
• Perpendicularity Condition (Dot Product): 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: If $A(x_A, y_A, z_A)$ is the midpoint of a segment connecting $P(x_P, y_P, z_P)$ and $P'(x_{P'}, y_{P'}, z_{P'})$, then $x_A = \frac{x_P+x_{P'}}{2}$, $y_A = \frac{y_P+y_{P'}}{2}$, $z_A = \frac{z_P+z_{P'}}{2}$. This can be rearranged to find $P'$: $P' = 2A - P$.
• Distance Formula in 3D: The square of the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $d^2 = (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$.

2. Step-by-Step Solution

Let the given point be $P(6,1,5)$. The given line $L$ is $\frac{x-1}{3}=\frac{y}{2}=\frac{z-2}{4}$. We need to find the square of the distance of the image of $P$ (let's call it $P'$) from the origin $O(0,0,0)$.

Step 1: Represent the Line Parametrically and Identify Key Vectors.

• What we are doing: We are converting the symmetric form of the line into a parametric form to easily represent any point on the line using a single variable $\lambda$. We also identify the direction vector of the line.
• Why we are doing this: The parametric form simplifies calculations involving points on the line, especially when applying conditions like perpendicularity. Let $\frac{x-1}{3}=\frac{y}{2}=\frac{z-2}{4} = \lambda$. Then, any general point $A$ on the line $L$ can be expressed as: $A(x,y,z) = (3\lambda+1, 2\lambda, 4\lambda+2)$ The direction vector of the line $L$ is obtained from the denominators of its symmetric form: $\vec{d} = 3\hat{i} + 2\hat{j} + 4\hat{k}$

Step 2: Find the Foot of the Perpendicular (Point A).

• What we are doing: We are finding the specific coordinates of point $A$ on line $L$ such that the line segment $PA$ is perpendicular to $L$.
• Why we are doing this: Point $A$ is crucial because it acts as the midpoint between the original point $P$ and its image $P'$. The perpendicularity condition allows us to determine the unique value of $\lambda$ for $A$. First, form the vector $\vec{PA}$ connecting point $P(6,1,5)$ to the general point $A(3\lambda+1, 2\lambda, 4\lambda+2)$ on the line: $\vec{PA} = A - P = ((3\lambda+1)-6)\hat{i} + (2\lambda-1)\hat{j} + ((4\lambda+2)-5)\hat{k}$ $\vec{PA} = (3\lambda-5)\hat{i} + (2\lambda-1)\hat{j} + (4\lambda-3)\hat{k}$ Since $PA$ is perpendicular to the line $L$, the vector $\vec{PA}$ must be perpendicular to the direction vector $\vec{d}$ of the line. Their dot product must be zero: $\vec{PA} \cdot \vec{d} = 0$ $(3\lambda-5)(3) + (2\lambda-1)(2) + (4\lambda-3)(4) = 0$ Expand and simplify the equation: $(9\lambda - 15) + (4\lambda - 2) + (16\lambda - 12) = 0$ $(9+4+16)\lambda + (-15-2-12) = 0$ $29\lambda - 29 = 0$ Solving for $\lambda$: $29\lambda = 29 \Rightarrow \lambda = 1$ Now, substitute $\lambda=1$ back into the coordinates of point $A$ to find its specific location: $A(3(1)+1, 2(1), 4(1)+2) = A(4, 2, 6)$ So, the foot of the perpendicular from $P$ to line $L$ is $A(4,2,6)$.

Step 3: Find the Image Point (Point P').

• What we are doing: We are using the property that the foot of the perpendicular, $A$, is the midpoint of the segment connecting the original point $P$ and its image $P'$.
• Why we are doing this: This is the fundamental geometric principle for finding the image of a point in a line. Let the image point be $P'(x', y', z')$. Using the midpoint formula, where $A$ is the midpoint of $PP'$: $A_x = \frac{P_x + P'_x}{2} \Rightarrow 4 = \frac{6 + x'}{2} \Rightarrow 8 = 6 + x' \Rightarrow x' = 2$ $A_y = \frac{P_y + P'_y}{2} \Rightarrow 2 = \frac{1 + y'}{2} \Rightarrow 4 = 1 + y' \Rightarrow y' = 3$ $A_z = \frac{P_z + P'_z}{2} \Rightarrow 6 = \frac{5 + z'}{2} \Rightarrow 12 = 5 + z' \Rightarrow z' = 7$ Alternatively, using the vector form $P' = 2A - P$: $P' = (2(4)-6, 2(2)-1, 2(6)-5) = (8-6, 4-1, 12-5) = (2,3,7)$ So, the image of the point $P(6,1,5)$ in the line $L$ is $P'(2,3,7)$.

Step 4: Calculate the Square of the Distance of P' from the Origin.

• What we are doing: We are calculating the square of the distance between the image point $P'(2,3,7)$ and the origin $O(0,0,0)$.
• Why we are doing this: This is the final requirement of the problem statement. Using the 3D distance formula for $O(0,0,0)$ and $P'(2,3,7)$: $OP'^2 = (2-0)^2 + (3-0)^2 + (7-0)^2$ $OP'^2 = 2^2 + 3^2 + 7^2$ $OP'^2 = 4 + 9 + 49$ $OP'^2 = 62$

3. Common Mistakes & Tips

• Common Mistake 1: Confusing Foot of Perpendicular with Image: A frequent error is to provide the coordinates or distance of the foot of the perpendicular ($A$) as the final answer. Remember, the problem specifically asks for the properties of the image point ($P'$).
• Common Mistake 2: Algebraic Errors: Carelessness in expanding dot products or applying the midpoint formula can lead to incorrect values of $\lambda$ or $P'$ coordinates. Double-check all calculations.
• Tip 1: Visualize: Always try to visualize the geometric setup. Understanding that the line acts as a mirror and the foot of the perpendicular is the midpoint helps solidify the method.
• Tip 2: Systematic Approach: Follow the steps methodically: parametric form, foot of perpendicular, image, final calculation. This minimizes errors and ensures all conditions are met.

4. Summary

To find the square of the distance of the image of a point from the origin, we followed a four-step process. First, we represented the given line in parametric form. Next, we determined the foot of the perpendicular from the original point to the line using the condition that the vector connecting them is orthogonal to the line's direction vector. With the foot of the perpendicular identified, we then found the coordinates of the image point by applying the midpoint formula. Finally, we calculated the square of the distance from this image point to the origin. The computed value for the square of the distance is 62.

5. Final Answer

The final answer is \boxed{62}.