MATCH THE FOLLOWING:
(A) Let image of the line $\frac{x-1}{3} = \frac{y-3}{5} = \frac{z-4}{-2}$ in the plane $2x - y + z + 3 = 0$ be $L$. A plane $7x + By + Cz + D = 0$ is such that it contains the line $L$ and perpendicular to the plane $2x - y + z + 3 = 0$ then value of $\frac{B+C+D}{10}$ is
Step-by-Step Solution
Key Concept: The required plane must contain the reflected line $L$ and be perpendicular to the original plane, creating two constraints that determine the plane uniquely.
First, find the image of the given line in the plane $2x - y + z + 3 = 0$. The line passes through $P(1,3,4)$ with direction vector $\vec{d}=(3,5,-2)$. Find the foot of perpendicular from $P$ to the plane: the perpendicular from $P$ has direction $(2,-1,1)$, giving foot $F$. The image point $P'$ is the reflection of $P$ across $F$. Next, any point on the original line reflects to a point on line $L$. The plane $7x + By + Cz + D = 0$ must contain $L$ and be perpendicular to $2x - y + z + 3 = 0$. Since the plane is perpendicular to the given plane, their normals satisfy: $(7, B, C) \cdot (2, -1, 1) = 0$, giving $14 - B + C = 0$, so $B = 14 + C$. Using that the plane contains points on the reflected line and solving the system of conditions yields specific values for $B$, $C$, and $D$, ultimately giving $\frac{B+C+D}{10}$ a numerical answer.
Correct Answer: I need to solve this step-by-step to find the image of the line and then determine the plane coefficients.
**Step 1: Find the foot of perpendicular from P(1,3,4) to plane 2x - y + z + 3 = 0**
The perpendicular from P has direction vector (2,-1,1