If the distance between the plane $Ax - 2y + z = d$ and the plane containing the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}$ is $\sqrt{6}$, then $|d|$ is equal to __________.
Step-by-Step Solution
Key Concept: For a plane to contain two lines, the normal vector must be perpendicular to both direction vectors.
Let the plane equation be $a(x-1) + b(y-2) + c(z-3) = 0$ containing two given lines. For each line to lie in the plane, we need $2a + 3b + 4c = 0$ and $3a + 4b + 5c = 0$. Solving these gives $\frac{a}{1} = \frac{b}{-2} = \frac{c}{1}$. From the first equation with these ratios: $(x-1) - 2(y-2) + (z-3) = 0$, which simplifies to $x - 2y + z = 0$.
Correct Answer: I need to find |d| given that the distance between the plane $Ax - 2y + z = d$ and the plane $x - 2y + z = 0$ is $\sqrt{6}$.
From the solution, weve established that the plane containing both lines is:
$$x - 2y + z = 0$$
The