Key Concepts and Formulas

1. Parametric Form of a Line: Any point on a line in symmetric form $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$ can be represented parametrically as $(x_0 + a\lambda, y_0 + b\lambda, z_0 + c\lambda)$, where $\lambda$ is a scalar parameter.
2. Foot of the Perpendicular: The foot of the perpendicular from a point P to a line L is a point P' on L such that the vector $\vec{PP'}$ is perpendicular to the direction vector of L. This condition is expressed by their dot product being zero: $\vec{PP'} \cdot \vec{d} = 0$.
3. Image of a Point in a Line: If Q is the image of point P in line L, and P' is the foot of the perpendicular from P to L, then P' is the midpoint of the line segment PQ. This implies $P' = \frac{P+Q}{2}$, or $Q = 2P' - P$.
4. Section Formula (Internal Division): For a point R dividing a line segment PQ internally in the ratio $m:n$, its coordinates are given by $R = \frac{n P + m Q}{m + n}$.

Step-by-Step Solution

Step 1: Parameterize the line L and find the vector $\vec{PP'}$

Let the given point be $P(1, 2, 3)$. The equation of the line L is given in symmetric form: $L: \frac{x - 6}{3} = \frac{y - 1}{2} = \frac{z - 2}{3}$ To find the foot of the perpendicular from P to L, let $P'$ be an arbitrary point on the line L. We can express its coordinates in terms of a parameter $\lambda$: $\frac{x - 6}{3} = \frac{y - 1}{2} = \frac{z - 2}{3} = \lambda$ So, the coordinates of $P'$ are $(3\lambda + 6, 2\lambda + 1, 3\lambda + 2)$.

Now, we form the vector $\vec{PP'}$ connecting point P to point P'. $\vec{PP'} = \langle (3\lambda + 6) - 1, (2\lambda + 1) - 2, (3\lambda + 2) - 3 \rangle$ $\vec{PP'} = \langle 3\lambda + 5, 2\lambda - 1, 3\lambda - 1 \rangle$

Step 2: Determine $\lambda$ using the perpendicularity condition and find P'

The direction vector of the line L is obtained from the denominators of its symmetric form: $\vec{d} = \langle 3, 2, 3 \rangle$. Since $P'$ is the foot of the perpendicular, the vector $\vec{PP'}$ must be perpendicular to the direction vector of the line L. The dot product of two perpendicular vectors is zero: $\vec{PP'} \cdot \vec{d} = 0$ $(3\lambda + 5)(3) + (2\lambda - 1)(2) + (3\lambda - 1)(3) = 0$ $9\lambda + 15 + 4\lambda - 2 + 9\lambda - 3 = 0$ Combine like terms: $(9\lambda + 4\lambda + 9\lambda) + (15 - 2 - 3) = 0$ $22\lambda + 10 = 0$ $22\lambda = -10$ $\lambda = -\frac{10}{22} = -\frac{5}{11}$

Now, substitute this value of $\lambda$ back into the parametric coordinates of P' to find the exact coordinates of the foot of the perpendicular: $x_{P'} = 3\left(-\frac{5}{11}\right) + 6 = -\frac{15}{11} + \frac{66}{11} = \frac{51}{11}$ $y_{P'} = 2\left(-\frac{5}{11}\right) + 1 = -\frac{10}{11} + \frac{11}{11} = \frac{1}{11}$ $z_{P'} = 3\left(-\frac{5}{11}\right) + 2 = -\frac{15}{11} + \frac{22}{11} = \frac{7}{11}$ Thus, the foot of the perpendicular is $P'\left(\frac{51}{11}, \frac{1}{11}, \frac{7}{11}\right)$.

Step 3: Establish the relationship between P, P', Q, and R

Let $P(x_P, y_P, z_P)$ be the original point. Let $Q(x_Q, y_Q, z_Q)$ be the image of P in the line L. Let $P'(x_{P'}, y_{P'}, z_{P'})$ be the foot of the perpendicular from P to L.

By the definition of an image point, the foot of the perpendicular P' is the midpoint of the line segment PQ. So, $P' = \frac{P + Q}{2}$. From this, we can express the coordinates of Q in terms of P and P': $Q = 2P' - P$

We are given that point $R(\alpha, \beta, \gamma)$ divides the line segment PQ internally in the ratio 1:3. Using the section formula, with $m=1$ and $n=3$: $R = \frac{3P + 1Q}{1 + 3} = \frac{3P + Q}{4}$ Now, substitute the expression for Q ($Q = 2P' - P$) into the formula for R: $R = \frac{3P + (2P' - P)}{4}$ $R = \frac{2P + 2P'}{4}$ $R = \frac{P + P'}{2}$ This is a significant simplification: R is the midpoint of the line segment PP'. This shortcut saves considerable calculation.

Step 4: Calculate the coordinates of R($\alpha, \beta, \gamma$)

Using the midpoint formula for P and P': $P(1, 2, 3)$ $P'\left(\frac{51}{11}, \frac{1}{11}, \frac{7}{11}\right)$ $R(\alpha, \beta, \gamma) = \left( \frac{x_P + x_{P'}}{2}, \frac{y_P + y_{P'}}{2}, \frac{z_P + z_{P'}}{2} \right)$ $\alpha = \frac{1 + \frac{51}{11}}{2} = \frac{\frac{11 + 51}{11}}{2} = \frac{\frac{62}{11}}{2} = \frac{62}{22} = \frac{31}{11}$ $\beta = \frac{2 + \frac{1}{11}}{2} = \frac{\frac{22 + 1}{11}}{2} = \frac{\frac{23}{11}}{2} = \frac{23}{22}$ $\gamma = \frac{3 + \frac{7}{11}}{2} = \frac{\frac{33 + 7}{11}}{2} = \frac{\frac{40}{11}}{2} = \frac{40}{22} = \frac{20}{11}$ So, the coordinates of R are $\left(\frac{31}{11}, \frac{23}{22}, \frac{20}{11}\right)$.

Step 5: Calculate the final value of $22(\alpha + \beta + \gamma)$

First, sum the coordinates of R: $\alpha + \beta + \gamma = \frac{31}{11} + \frac{23}{22} + \frac{20}{11}$ To add these fractions, we use a common denominator of 22: $\alpha + \beta + \gamma = \frac{31 \times 2}{22} + \frac{23}{22} + \frac{20 \times 2}{22}$ $\alpha + \beta + \gamma = \frac{62}{22} + \frac{23}{22} + \frac{40}{22}$ $\alpha + \beta + \gamma = \frac{62 + 23 + 40}{22} = \frac{125}{22}$ Finally, multiply this sum by 22: $22(\alpha + \beta + \gamma) = 22 \times \left(\frac{125}{22}\right)$ $22(\alpha + \beta + \gamma) = 125$

Common Mistakes & Tips

• Arithmetic with Fractions: Be extremely careful when adding, subtracting, and dividing fractions, especially when dealing with common denominators.
• Geometric Interpretation: Always visualize or sketch the points and line. Understanding that P' is the midpoint of PQ, and then deriving that R is the midpoint of PP', is a crucial time-saving shortcut.
• Dot Product Errors: Ensure the dot product is correctly applied for perpendicularity. The vector connecting the point to the line must be perpendicular to the line's direction vector.
• Image vs. Foot: Do not confuse the image of a point (Q) with the foot of the perpendicular (P'). The foot of the perpendicular is the midpoint of the segment connecting the point to its image.

Summary

We began by finding the foot of the perpendicular P' from point P to the given line L using the parametric form of the line and the dot product condition for perpendicularity. After calculating P', we utilized the geometric relationship that P' is the midpoint of PQ (where Q is the image of P). By substituting this into the section formula for R dividing PQ in the ratio 1:3, we discovered the elegant shortcut that R is simply the midpoint of PP'. We then calculated the coordinates of R and summed them, finally multiplying by 22 to arrive at the solution.

The final answer is $\boxed{125}$.