Step 1: Understanding the Question:
We need to find the mathematical equations of two planes that are parallel to the given reference plane $x - 2y + 2z + 4 = 0$, such that their perpendicular distance to the point $(1, 2, 3)$ is exactly 1 unit.
Step 2: Key Formula or Approach:
1. Any plane parallel to $Ax + By + Cz + D = 0$ shares identical direction normal coefficients and can be written as:
$$Ax + By + Cz + \lambda = 0$$
2. The formula for the perpendicular distance $d$ from a point $(x_1, y_1, z_1)$ to a plane is:
$$d = \frac{|Ax_1 + By_1 + Cz_1 + \lambda|}{\sqrt{A^2 + B^2 + C^2}}$$
Step 3: Detailed Explanation:
3. Write down the expression for a parallel plane template:
$$x - 2y + 2z + \lambda = 0$$
4. Substitute the given point $(1, 2, 3)$ and set distance $d = 1$:
$$1 = \frac{|1(1) - 2(2) + 2(3) + \lambda|}{\sqrt{1^2 + (-2)^2 + 2^2}}$$
$$1 = \frac{|1 - 4 + 6 + \lambda|}{\sqrt{1 + 4 + 4}}$$
$$1 = \frac{|3 + \lambda|}{\sqrt{9}} = \frac{|3 + \lambda|}{3}$$
5. Clear the denominator by multiplying across:
$$|3 + \lambda| = 3$$
6. Solve the absolute value equation, which splits into two cases:
Case 1: $3 + \lambda = 3 \implies \lambda = 0$
Case 2: $3 + \lambda = -3 \implies \lambda = -6$
5. Plug these values back into our parallel plane template:
For $\lambda = 0$: $x - 2y + 2z = 0$
For $\lambda = -6$: $x - 2y + 2z - 6 = 0$
Step 4: Final Answer:
The equations of the parallel planes correspond to option (B).