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Step-by-Step Solution
Step 1: Express the Line L in Parametric Form
The given line L is:
\frac{x - 6}{3} \;=\; \frac{y - 1}{2} \;=\; \frac{z - 2}{3} \;=\; \lambda
We can write its parametric equations as:
x = 6 + 3\lambda,\quad y = 1 + 2\lambda,\quad z = 2 + 3\lambda.
Step 2: Identify the Given Point and Recall Reflection Relation
We have a point P(1,\,2,\,3) and its reflection Q in the given line L.
The foot of the perpendicular from P on the line L is denoted by P' , and for a reflection across a line, P' is the midpoint of the segment PQ.
Hence,
P' \;=\; \frac{P + Q}{2}.
Step 3: Determine the Foot of the Perpendicular P'
Let P' = (3\lambda + 6, \; 2\lambda + 1, \; 3\lambda + 2) be a general point on the line (from Step 1).
Since P' is the foot of the perpendicular from P , the vector
\overrightarrow{PP'} = \bigl((3\lambda+6) - 1,\,(2\lambda+1) - 2,\,(3\lambda+2) - 3\bigr)
should be perpendicular to the direction vector of the line L, which is (3,\,2,\,3).
Thus,
\overrightarrow{PP'} \cdot \text{(direction of L)} = 0.
Substituting, we get:
\bigl(3\lambda + 5, \, 2\lambda - 1, \, 3\lambda - 1\bigr) \,\cdot\, (3,\,2,\,3) = 0.
So,
(3\lambda+5)\cdot 3 \;+\; (2\lambda - 1)\cdot 2 \;+\; (3\lambda-1)\cdot 3 \;=\; 0.
Simplifying,
9\lambda + 15 \;+\; 4\lambda - 2 \;+\; 9\lambda - 3 \;=\; 0,
22\lambda + (15 - 2 - 3) \;=\; 0,
22\lambda + 10 \;=\; 0 \;\;\;\; \Longrightarrow \;\;\;\; \lambda = -\frac{10}{22} = -\frac{5}{11}.
Step 4: Find Coordinates of the Foot P'
Substitute \lambda = -\tfrac{5}{11} back into the parametric form:
P' \bigl(3\lambda + 6, \; 2\lambda + 1, \; 3\lambda + 2\bigr).
So,
P'_x = 3\Bigl(-\frac{5}{11}\Bigr) + 6 = -\frac{15}{11} + \frac{66}{11} = \frac{51}{11},
P'_y = 2\Bigl(-\frac{5}{11}\Bigr) + 1 = -\frac{10}{11} + \frac{11}{11} = \frac{1}{11},
P'_z = 3\Bigl(-\frac{5}{11}\Bigr) + 2 = -\frac{15}{11} + \frac{22}{11} = \frac{7}{11}.
Hence,
P' = \Bigl(\frac{51}{11}, \,\frac{1}{11}, \,\frac{7}{11}\Bigr).
Step 5: The Required Point R That Divides PQ in the Ratio 1:3
By definition of reflection, P' is the midpoint of PQ.
Hence,
P' = \frac{P + Q}{2}.
The question states: R divides internally the line segment PQ in the ratio 1:3.
From the provided solution steps, it is established (using specific geometry or direct algebraic relations) that this point R is effectively the midpoint of P and P' .
Therefore,
R(\alpha,\beta,\gamma) = \frac{P + P'}{2}.
Let us calculate this midpoint of P(1,2,3) and P'\bigl(\tfrac{51}{11},\,\tfrac{1}{11},\,\tfrac{7}{11}\bigr) :
\alpha = \frac{1 + \frac{51}{11}}{2} = \frac{\frac{11}{11} + \frac{51}{11}}{2} = \frac{62}{11} \cdot \frac{1}{2} = \frac{62}{22},
\beta = \frac{2 + \frac{1}{11}}{2} = \frac{\frac{22}{11} + \frac{1}{11}}{2} = \frac{23}{11} \cdot \frac{1}{2} = \frac{23}{22},
\gamma = \frac{3 + \frac{7}{11}}{2} = \frac{\frac{33}{11} + \frac{7}{11}}{2} = \frac{40}{11} \cdot \frac{1}{2} = \frac{40}{22}.
Step 6: Compute the Required Expression 22(\alpha + \beta + \gamma)
We have
\alpha = \frac{62}{22}, \quad \beta = \frac{23}{22}, \quad \gamma = \frac{40}{22}.
So,
\alpha + \beta + \gamma = \frac{62}{22} + \frac{23}{22} + \frac{40}{22} = \frac{62 + 23 + 40}{22} = \frac{125}{22}.
Therefore,
22(\alpha + \beta + \gamma) = 22 \times \frac{125}{22} = 125.
Answer
22(\alpha + \beta + \gamma) = 125.
