Question

Let \(P(\alpha, \beta, \gamma)\) be the image of the point \(Q(1, 6, 4)\) in the line \[ \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}. \] Then \(2\alpha + \beta + \gamma\) is equal to _______.

Answer 1

Correct Answer: 11

Given the line equation:
\[ \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} \]

Step 1: Direction vector of the line
\[ \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k} \]

Step 2: Coordinates of point A
\[ A(t, 2t + 1, 3t + 2) \]

Step 3: Position vector of point Q
\[ Q(1, 5, 2) \] Therefore,
\[ \overrightarrow{QA} = (t - 1)\hat{i} + (2t - 5)\hat{j} + (3t - 2)\hat{k} \]

Step 4: Perpendicular condition
Since \( \overrightarrow{QA} \cdot \vec{b} = 0 \), we have:
\[ (t - 1) + 2(2t - 5) + 3(3t - 2) = 0 \] Simplifying:
\[ t - 1 + 4t - 10 + 9t - 6 = 0 \] \[ 14t = 17 \] \[ t = \frac{17}{14} \]

Step 5: Substituting \( t = \frac{17}{14} \)
\[ \alpha = t = \frac{20}{14}, \quad \beta = \frac{12}{14}, \quad \gamma = \frac{102}{14} \]

Step 6: Required Expression
\[ 2\alpha + \beta + \gamma = \frac{154}{14} = 11 \]

Final Answer:
\[ \boxed{11} \]

Answer 2

Let the line be represented by:
\[x = t, \quad y = 2t + 1, \quad z = 3t + 2.\]
Given the point \(Q(1, 6, 4)\), we find the foot of the perpendicular \(A\) by letting:
\[A\left(\frac{17}{14}, \frac{48}{14}, \frac{79}{14}\right).\]
The direction vector of the line is:
\[\mathbf{b} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}.\]
To find the image point \(P(\alpha, \beta, \gamma)\), we reflect \(Q\) across \(A\) using:
\[\alpha = \frac{20}{14}, \quad \beta = \frac{12}{14}, \quad \gamma = \frac{102}{14}.\]
Calculating \(2\alpha + \beta + \gamma\):
\[2\alpha + \beta + \gamma = \frac{154}{14} = 11.\]
Answer: 11.