The image of the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{4}$ in the plane $x + 2y + z = 12$ is:
$\frac{x-6}{2} = \frac{y+\frac{7}{2}}{-2} = \frac{z-13}{2}$
$\frac{x-6}{4} = \frac{y+\frac{7}{2}}{-7} = \frac{z-13}{10}$
$\frac{x+6}{2} = \frac{y-\frac{7}{2}}{-3} = \frac{z-13}{6}$
None of these
Step-by-Step Solution
Key Concept: Image of a point in a plane is found by reflecting across the plane using the condition that the midpoint lies on the plane and the joining line is perpendicular to it.
Any point on the given line is $(2r+1, -r-1, 4r+3)$. If this point lies on the plane, substitute into the plane equation: $2r+1-2r-2+4r+3=12$, which gives $r=\frac{5}{2}$. The point of intersection is $\left(6, -\frac{7}{2}, 13\right)$. To find the image of $(1,-1,3)$, use the condition that $(1,-1,3)$ and its image are symmetric about the plane, with their midpoint lying on the plane. This yields $\lambda=\frac{10}{3}$ and the image is $\left(\frac{8}{3}, \frac{7}{3}, \frac{14}{3}\right)$. The required line equation is $\frac{x-6}{10/3}=\frac{y+7/2}{-55/6}=\frac{z-13}{25/3}$ or $\frac{x-6}{4}=\frac{y+7/2}{-7}=\frac{z-13}{10}$.
Correct Answer: 2