The equations of the planes through the origin which are parallel to the line $\frac{x - 1}{2} = \frac{y + 3}{-1} = \frac{z + 1}{-2}$ and at a distance $5/3$ from it are:
Step-by-Step Solution
Key Concept: Use the parallel condition (normal perpendicular to direction) and distance formula to set up a system of equations.
Let the plane equation be $Ax + By + Cz = 0$. Since the line is parallel to the plane, $2A - B - 2C = 0$. The distance formula with the given point $(1, -3, -1)$ gives $|A - 3B - C| = \frac{5}{3}\sqrt{A^2 + B^2 + C^2}$. Solving these three conditions simultaneously yields the plane coefficients.
Correct Answer: I need to find planes through the origin that are parallel to the given line and at distance 5/3 from it.
**Setup:**
- Plane equation: $Ax + By + Cz = 0$ (passes through origin)
- Line: $\frac{x-1}{2} = \frac{y+3}{-1