A plane which bisects the angle between the two gives planes `2x - y +2z - 4 = 0` and `x +2y +2z - 2 = 0`, passes through the point:

Options

A. `(1,-4,1)`
B. `(1,4,-1)`
C. `(2,4,1)`
D. `(2,-4,1)`

Answer

To find the plane that bisects the angle between the two given planes \(2x - y + 2z - 4 = 0\) and \(x + 2y + 2z - 2 = 0\), we can use the formula for the angle bisector of two planes.

### Step-by-Step Solution:

1. **Identify the coefficients of the given planes**:
- For the first plane \(P_1: 2x - y + 2z - 4 = 0\), the coefficients are:
- \(A_1 = 2\), \(B_1 = -1\), \(C_1 = 2\), \(D_1 = -4\)
- For the second plane \(P_2: x + 2y + 2z - 2 = 0\), the coefficients are:
- \(A_2 = 1\), \(B_2 = 2\), \(C_2 = 2\), \(D_2 = -2\)

2. **Write the equation for the angle bisector**:
The equation of the angle bisector of two planes can be expressed as:
\[ \frac{A_1x + B_1y + C_1z + D_1}{\sqrt{A_1^2 + B_1^2 + C_1^2}} = \pm \frac{A_2x + B_2y + C_2z + D_2}{\sqrt{A_2^2 + B_2^2 + C_2^2}} \]

3. **Calculate the denominators**:
- For the first plane:
\[ \sqrt{A_1^2 + B_1^2 + C_1^2} = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]
- For the second plane:
\[ \sqrt{A_2^2 + B_2^2 + C_2^2} = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

4. **Substitute into the angle bisector equation**:
\[ \frac{2x - y + 2z - 4}{3} = \pm \frac{x + 2y + 2z - 2}{3} \]
This simplifies to:
\[ 2x - y + 2z - 4 = \pm (x + 2y + 2z - 2) \]

5. **Separate into two cases**:
- **Case 1** (using the positive sign):
\[ 2x - y + 2z - 4 = x + 2y + 2z - 2 \]
Rearranging gives:
\[ 2x - x - y - 2y + 2z - 2z - 4 + 2 = 0 \implies x - 3y - 2 = 0 \implies x - 3y + 2 = 0 \]
- **Case 2** (using the negative sign):
\[ 2x - y + 2z - 4 = - (x + 2y + 2z - 2) \]
Rearranging gives:
\[ 2x - y + 2z - 4 + x + 2y + 2z - 2 = 0 \implies 3x + y + 4z - 6 = 0 \]

6. **Final equations of the bisectors**:
- From Case 1: \(x - 3y + 2 = 0\)
- From Case 2: \(3x + y + 4z - 6 = 0\)

7. **Check if a point lies on the bisector**:
Let's check the point \((2, -4, 1)\):
- For \(x - 3y + 2 = 0\):
\[ 2 - 3(-4) + 2 = 2 + 12 + 2 = 16 \quad (\text{not satisfied}) \]
- For \(3x + y + 4z - 6 = 0\):
\[ 3(2) + (-4) + 4(1) - 6 = 6 - 4 + 4 - 6 = 0 \quad (\text{satisfied}) \]

Thus, the plane that bisects the angle between the two given planes passes through the point \((2, -4, 1)\).