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

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.