The two points`(1,1,1) and (-3,0,1)` with respect to the plane `3x+4y-12z+13=0` lie on

Options

A. opposite side
B. same side
C. on the plane
D. None of these

Answer

To determine the position of the points \( (1, 1, 1) \) and \( (-3, 0, 1) \) with respect to the plane defined by the equation \( 3x + 4y - 12z + 13 = 0 \), we will calculate the perpendicular distance of each point from the plane.

### Step-by-Step Solution:

1. **Identify the coefficients of the plane equation**:
The plane equation is given as \( 3x + 4y - 12z + 13 = 0 \). Here, the coefficients are:
- \( a = 3 \)
- \( b = 4 \)
- \( c = -12 \)
- \( d = 13 \)

2. **Use the distance formula from a point to a plane**:
The distance \( D \) from a point \( (x_1, y_1, z_1) \) to the plane \( ax + by + cz + d = 0 \) is given by:
\[ D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \]

3. **Calculate the distance for the first point \( (1, 1, 1) \)**:
- Substitute \( (x_1, y_1, z_1) = (1, 1, 1) \) into the formula:
\[ D_1 = \frac{|3(1) + 4(1) - 12(1) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} \]
- Calculate the numerator:
\[ = |3 + 4 - 12 + 13| = |8| = 8 \]
- Calculate the denominator:
\[ = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 \]
- Therefore, the distance \( D_1 \) is:
\[ D_1 = \frac{8}{13} \]

4. **Calculate the distance for the second point \( (-3, 0, 1) \)**:
- Substitute \( (x_1, y_1, z_1) = (-3, 0, 1) \) into the formula:
\[ D_2 = \frac{|3(-3) + 4(0) - 12(1) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} \]
- Calculate the numerator:
\[ = |-9 + 0 - 12 + 13| = |-8| = 8 \]
- The denominator remains the same:
\[ = 13 \]
- Therefore, the distance \( D_2 \) is:
\[ D_2 = \frac{8}{13} \]

5. **Conclusion**:
Since both distances \( D_1 \) and \( D_2 \) are equal, it indicates that both points \( (1, 1, 1) \) and \( (-3, 0, 1) \) lie on the same side of the plane.

### Final Answer:
Both points \( (1, 1, 1) \) and \( (-3, 0, 1) \) lie on the same side of the plane \( 3x + 4y - 12z + 13 = 0 \).