This problem requires finding the mirror image of a point with respect to a given plane, a fundamental concept in 3D geometry. We'll use the standard formula for the image of a point and then perform the required summation.

1. Key Concepts and Formulas

• Image of a Point with Respect to a Plane: Given a point $P(x_1, y_1, z_1)$ and a plane $Ax + By + Cz + D = 0$, its mirror image $P'(\alpha, \beta, \gamma)$ is found using the formula: $\frac{\alpha - x_1}{A} = \frac{\beta - y_1}{B} = \frac{\gamma - z_1}{C} = -2 \frac{Ax_1 + By_1 + Cz_1 + D}{A^2 + B^2 + C^2}$
• Geometric Interpretation: This formula is derived from two geometric properties:
  1. The line segment connecting the original point $P$ and its image $P'$ is perpendicular to the plane. The direction ratios of this line $(A, B, C)$ are proportional to the normal vector of the plane.
  2. The midpoint of the line segment $PP'$ lies on the plane. The factor of $-2$ ensures that the image point is located symmetrically on the opposite side of the plane from the original point.

2. Step-by-Step Solution

Step 1: Identify Given Information

First, we extract the coordinates of the given point and the coefficients of the plane equation.

• The given point $P(x_1, y_1, z_1)$ is $(1, 3, 5)$. So, $x_1 = 1$, $y_1 = 3$, $z_1 = 5$.
• The equation of the plane is $4x - 5y + 2z = 8$. To use the formula, we rewrite it in the standard form $Ax + By + Cz + D = 0$. Thus, the plane equation is $4x - 5y + 2z - 8 = 0$.
• Comparing this with $Ax + By + Cz + D = 0$, we identify the coefficients: $A = 4$ $B = -5$ $C = 2$ $D = -8$
• The mirror image is denoted as $(\alpha, \beta, \gamma)$.

Step 2: Calculate the Numerator Term ($Ax_1 + By_1 + Cz_1 + D$)

This term represents the signed "value" of the plane equation when the coordinates of the point $P$ are substituted into it. It is crucial for determining the distance of the point from the plane and thus the displacement of the image.

Substitute the values of $A, B, C, D$ and $x_1, y_1, z_1$: $Ax_1 + By_1 + Cz_1 + D = 4(1) + (-5)(3) + 2(5) + (-8)$ $= 4 - 15 + 10 - 8$ $= 27$

Step 3: Calculate the Denominator Term ($A^2 + B^2 + C^2$)

This term represents the square of the magnitude of the normal vector to the plane. It normalizes the expression in the formula.

Substitute the values of $A, B, C$: $A^2 + B^2 + C^2 = 4^2 + (-5)^2 + 2^2$ $= 16 + 25 + 4$ $= 45$

Step 4: Determine the Common Ratio ($k$)

Now, we substitute the calculated numerator and denominator values into the formula to find the common ratio, which we'll call $k$. This ratio dictates the position of the image point.

$k = -2 \frac{Ax_1 + By_1 + Cz_1 + D}{A^2 + B^2 + C^2}$ $k = -2 \frac{27}{45}$ Simplify the fraction $\frac{27}{45}$ by dividing both numerator and denominator by 9: $\frac{27 \div 9}{45 \div 9} = \frac{3}{5}$. $k = -2 \left( \frac{3}{5} \right)$ $k = -\frac{6}{5}$

So, our formula becomes: $\frac{\alpha - 1}{4} = \frac{\beta - 3}{-5} = \frac{\gamma - 5}{2} = -\frac{6}{5}$

Step 5: Determine the Coordinates of the Image Point $(\alpha, \beta, \gamma)$

Equate each part of the ratio to $k = -\frac{6}{5}$ to find the individual coordinates of the image point.

• For $\alpha$: $\frac{\alpha - 1}{4} = -\frac{6}{5}$ $\alpha - 1 = 4 \times \left(-\frac{6}{5}\right)$ $\alpha - 1 = -\frac{24}{5}$ $\alpha = 1 - \frac{24}{5} = \frac{5}{5} - \frac{24}{5} = -\frac{19}{5}$

• For $\beta$: $\frac{\beta - 3}{-5} = -\frac{6}{5}$ $\beta - 3 = -5 \times \left(-\frac{6}{5}\right)$ $\beta - 3 = 6$ $\beta = 3 + 6 = 9$

• For $\gamma$: $\frac{\gamma - 5}{2} = -\frac{6}{5}$ $\gamma - 5 = 2 \times \left(-\frac{6}{5}\right)$ $\gamma - 5 = -\frac{12}{5}$ $\gamma = 5 - \frac{12}{5} = \frac{25}{5} - \frac{12}{5} = \frac{13}{5}$

Thus, the mirror image point $(\alpha, \beta, \gamma)$ is $\left(-\frac{19}{5}, 9, \frac{13}{5}\right)$.

Step 6: Calculate $5(\alpha + \beta + \gamma)$

The problem asks for the value of $5(\alpha + \beta + \gamma)$. First, sum the coordinates: $\alpha + \beta + \gamma = -\frac{19}{5} + 9 + \frac{13}{5}$ Convert $9$ to a fraction with a denominator of $5$: $9 = \frac{45}{5}$. $\alpha + \beta + \gamma = -\frac{19}{5} + \frac{45}{5} + \frac{13}{5}$ $\alpha + \beta + \gamma = \frac{-19 + 45 + 13}{5}$ $\alpha + \beta + \gamma = \frac{26 + 13}{5}$ $\alpha + \beta + \gamma = \frac{39}{5}$

Finally, multiply this sum by 5: $5(\alpha + \beta + \gamma) = 5 \times \frac{39}{5}$ $5(\alpha + \beta + \gamma) = 39$

3. Common Mistakes & Tips

• Sign of D: Always ensure the plane equation is in the form $Ax + By + Cz + D = 0$. If given as $Ax + By + Cz = d$, then $D = -d$. A common error is using the incorrect sign for $D$.
• The Factor of -2: Do not forget the $-2$ in the image formula. Using $-1$ gives the foot of the perpendicular, and omitting the negative sign (i.e., using $2$) would result in a point on the same side of the plane as the original point.
• Arithmetic Accuracy: These problems involve several arithmetic steps, often with fractions and negative numbers. Double-check all calculations to avoid errors.
• Understanding the Derivation: While memorizing the formula is helpful, understanding its derivation (via the perpendicular line and midpoint conditions) can prevent errors and aid in problem-solving if the exact formula is forgotten.

4. Summary

To find the mirror image of a point $P(x_1, y_1, z_1)$ with respect to a plane $Ax + By + Cz + D = 0$, we use the formula $\frac{\alpha - x_1}{A} = \frac{\beta - y_1}{B} = \frac{\gamma - z_1}{C} = -2 \frac{Ax_1 + By_1 + Cz_1 + D}{A^2 + B^2 + C^2}$. For point $(1, 3, 5)$ and plane $4x - 5y + 2z - 8 = 0$, we identified $x_1=1, y_1=3, z_1=5$ and $A=4, B=-5, C=2, D=-8$. We calculated $Ax_1 + By_1 + Cz_1 + D = 27$ and $A^2 + B^2 + C^2 = 45$. This led to a common ratio $k = -2 \times \frac{27}{45} = -\frac{6}{5}$. Using this ratio, we found the image point $(\alpha, \beta, \gamma) = \left(-\frac{19}{5}, 9, \frac{13}{5}\right)$. Finally, we calculated $5(\alpha + \beta + \gamma) = 5 \left(-\frac{19}{5} + 9 + \frac{13}{5}\right) = 5 \left(\frac{39}{5}\right) = 39$.

The final answer is $\boxed{\text{39}}$, which corresponds to option (A).