Step 1: Understanding the Concept:
The locus of points equidistant from two intersecting planes consists of two planes that bisect the angles between the original planes. We need to find the equations of these bisecting planes and then check which of the given points lies on one of them.
Step 2: Key Formula or Approach:
The equations of the planes that bisect the angles between $A_1x+B_1y+C_1z+D_1=0$ and $A_2x+B_2y+C_2z+D_2=0$ are given by:
\[ \frac{A_1x+B_1y+C_1z+D_1}{\sqrt{A_1^2+B_1^2+C_1^2}} = \pm \frac{A_2x+B_2y+C_2z+D_2}{\sqrt{A_2^2+B_2^2+C_2^2}} \]
Step 3: Detailed Explanation:
Let the two planes be $P_1: x+2y+2z-2=0$ and $P_2: 2x-y+2z-4=0$.
First, calculate the magnitude of the normal vectors:
For $P_1$: $\sqrt{1^2+2^2+2^2} = \sqrt{1+4+4} = \sqrt{9} = 3$.
For $P_2$: $\sqrt{2^2+(-1)^2+2^2} = \sqrt{4+1+4} = \sqrt{9} = 3$.
The equations of the bisecting planes are:
\[ \frac{x+2y+2z-2}{3} = \pm \frac{2x-y+2z-4}{3} \]
This gives two equations:
Case 1 (with '+'):
$x+2y+2z-2 = 2x-y+2z-4$
$x - 3y - 2 = 0$
Case 2 (with '-'):
$x+2y+2z-2 = -(2x-y+2z-4)$
$x+2y+2z-2 = -2x+y-2z+4$
$3x+y+4z-6=0$
Now, we test the given points to see if they satisfy either equation.
Let's test point (B) (2,-4,1):
In the first bisecting plane: $1(2) - 3(-4) - 2 = 2 + 12 - 2 = 12 \ne 0$. The point is not on this plane.
In the second bisecting plane: $3(2) + (-4) + 4(1) - 6 = 6 - 4 + 4 - 6 = 0$. The point satisfies this equation.
Since the point (2,-4,1) lies on one of the angle bisecting planes, it is the correct answer.
Step 4: Final Answer:
The plane passes through the point $(2,-4,1)$.