Key Concepts and Formulas

1. Image of a Point in a Line: The image $P'$ of a point $P$ in a line $L$ is such that $L$ is the perpendicular bisector of the segment $PP'$. This implies two conditions:
   • Perpendicularity: The line segment $PQ$ (where $Q$ is the foot of the perpendicular from $P$ to $L$) is perpendicular to the line $L$. The dot product of their direction ratios (DRs) is zero.
   • Midpoint Property: The foot of the perpendicular $Q$ is the midpoint of the line segment connecting the original point $P$ and its image $P'$.
2. 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 $x=x_0+a\lambda$, $y=y_0+b\lambda$, $z=z_0+c\lambda$. Any point on the line can be represented using this parameter $\lambda$. The direction ratios of the line are $(a,b,c)$.
3. Direction Ratios of a Line Segment: For two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, the direction ratios of the line segment connecting them are $(x_2-x_1, y_2-y_1, z_2-z_1)$.
4. Midpoint Formula: The midpoint $M(x_m, y_m, z_m)$ of a segment connecting $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $M\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)$.

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Step-by-Step Solution

Step 1: Parametrize the Line and Define the Points

We begin by representing any general point on the given line using a parameter. This is essential for expressing the coordinates of the foot of the perpendicular in terms of a single variable.

The given line is: $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{5}$ Let's set each part equal to a parameter $\lambda$: $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{5} = \lambda$ From this, we can express the coordinates of a general point $Q$ on the line in terms of $\lambda$: $x_Q = 2\lambda+1$ $y_Q = 3\lambda-1$ $z_Q = 5\lambda+2$ So, the coordinates of a general point $Q$ on the line are $Q(2\lambda+1, 3\lambda-1, 5\lambda+2)$.

The given point is $P(8,5,7)$. The image point is $P'(\alpha, \beta, \gamma)$.

Why this step? Parametrizing the line allows us to represent the coordinates of the foot of the perpendicular $Q$ (which lies on the line) using a single variable $\lambda$. This simplifies calculations when we apply the perpendicularity condition.

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Step 2: Determine Direction Ratios of PQ

Next, we find the direction ratios (DRs) of the line segment $PQ$, where $Q$ is the foot of the perpendicular from $P$ to the line. The DRs of a line segment connecting $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are $(x_2-x_1, y_2-y_1, z_2-z_1)$.

The coordinates are $P(8,5,7)$ and $Q(2\lambda+1, 3\lambda-1, 5\lambda+2)$. The DRs of $PQ$ are: $( (2\lambda+1) - 8, (3\lambda-1) - 5, (5\lambda+2) - 7 )$ $= (2\lambda-7, 3\lambda-6, 5\lambda-5)$ The DRs of the given line $L$ are obtained directly from its symmetric form: $(2,3,5)$.

Why this step? We need the DRs of $PQ$ and the line $L$ to apply the perpendicularity condition in the next step, which will help us find the specific value of $\lambda$.

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Step 3: Apply Perpendicularity Condition to Find the Foot of the Perpendicular

Since $Q$ is the foot of the perpendicular from $P$ to the line $L$, the line segment $PQ$ must be perpendicular to the line $L$. Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if their dot product is zero: $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.

Using the DRs of $PQ$ $(2\lambda-7, 3\lambda-6, 5\lambda-5)$ and the DRs of line $L$ $(2,3,5)$: $2(2\lambda-7) + 3(3\lambda-6) + 5(5\lambda-5) = 0$ Let's expand and solve for $\lambda$: $(4\lambda - 14) + (9\lambda - 18) + (25\lambda - 25) = 0$ Combine like terms: $(4\lambda + 9\lambda + 25\lambda) + (-14 - 18 - 25) = 0$ $38\lambda - 60.8 = 0$ $38\lambda = 60.8$ $\lambda = \frac{60.8}{38} = 1.6$

Why this step? This is the core step to locate the specific point $Q$ on the line $L$ that is closest to $P$. The perpendicularity condition uniquely determines the value of $\lambda$ for the foot of the perpendicular.

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Step 4: Calculate the Coordinates of the Foot of the Perpendicular (Q)

Now that we have the value of $\lambda$, we can substitute it back into the parametric equations for $Q$ to find its exact coordinates.

Substitute $\lambda = 1.6$ into $Q(2\lambda+1, 3\lambda-1, 5\lambda+2)$: $x_Q = 2(1.6)+1 = 3.2+1 = 4.2$ $y_Q = 3(1.6)-1 = 4.8-1 = 3.8$ $z_Q = 5(1.6)+2 = 8+2 = 10$ So, the foot of the perpendicular is $Q(4.2, 3.8, 10)$.

Why this step? The foot of the perpendicular $Q$ is a critical intermediate point. It is the point on the line that is equidistant from $P$ and its image $P'$.

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Step 5: Use the Midpoint Formula to Find the Image Point

The foot of the perpendicular $Q$ is the midpoint of the original point $P(8,5,7)$ and its image $P'(\alpha, \beta, \gamma)$. The midpoint formula for two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $M\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)$.

Using $P(8,5,7)$, $P'(\alpha, \beta, \gamma)$, and $Q(4.2, 3.8, 10)$: For the x-coordinate: $4.2 = \frac{8+\alpha}{2}$ $8.4 = 8+\alpha \implies \alpha = 0.4$ For the y-coordinate: $3.8 = \frac{5+\beta}{2}$ $7.6 = 5+\beta \implies \beta = 2.6$ For the z-coordinate: $10 = \frac{7+\gamma}{2}$ $20 = 7+\gamma \implies \gamma = 13$ Thus, the image of the point $(8,5,7)$ in the given line is $P'(0.4, 2.6, 13)$.

Why this step? This is the final step to find the coordinates of the image. The midpoint property is a direct consequence of how reflection works: the line of reflection acts as the perpendicular bisector of the segment connecting a point and its image.

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Step 6: Calculate the Required Sum

The question asks for the sum $\alpha+\beta+\gamma$. $\alpha+\beta+\gamma = 0.4+2.6+13 = 3+13 = 16$

Why this step? This is the direct answer to the problem's specific request.

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Common Mistakes & Tips

• Parametrization: Ensure correct signs and coefficients when expressing general points on the line. For example, $y+1$ becomes $y = 3\lambda-1$.
• Direction Ratios: Be careful with the order of subtraction when calculating DRs for $PQ$.
• Algebraic Errors: The perpendicularity condition often leads to a linear equation in $\lambda$. Meticulously expand and combine terms to avoid errors.
• Midpoint Formula: Remember that the foot of the perpendicular $Q$ is the midpoint of $P$ and its image $P'$. It's a common mistake to use $Q$ as the image directly.
• Conceptual Understanding: Always visualize the geometry: $P$, $Q$, and $P'$ are collinear, with $Q$ in the middle, and $PQ$ is perpendicular to the line.

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Summary

To find the image of a point in a line, we first parametrize the line. Then, we find the foot of the perpendicular from the given point to the line using the perpendicularity condition (dot product of direction ratios is zero). Finally, we use the fact that this foot of the perpendicular is the midpoint of the original point and its image to calculate the image coordinates. For the point $(8,5,7)$ and the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{5}$, the image is $(0.4, 2.6, 13)$. The sum of its coordinates is $0.4+2.6+13=16$.

The final answer is $\boxed{\text{16}}$, which corresponds to option (A).