Find the equations of the planes that passes through three points :
(a ) (1,1-1) , (6,4,-5) (-4,-2,3)
(b) 1,1,0), (1,2,1) , (-2,2,-1).

Answer

To find the equations of the planes that pass through the given points, we will follow a systematic approach.

### Part (a): Points (1, 1, -1), (6, 4, -5), (-4, -2, 3)

1. **Define the Points as Vectors**:
Let the points be represented as vectors:
- \( A = (1, 1, -1) \)
- \( B = (6, 4, -5) \)
- \( C = (-4, -2, 3) \)

2. **Calculate Vectors \( \vec{B} - \vec{A} \) and \( \vec{C} - \vec{A} \)**:
- \( \vec{B} - \vec{A} = (6 - 1, 4 - 1, -5 + 1) = (5, 3, -4) \)
- \( \vec{C} - \vec{A} = (-4 - 1, -2 - 1, 3 + 1) = (-5, -3, 4) \)

3. **Cross Product of \( \vec{B} - \vec{A} \) and \( \vec{C} - \vec{A} \)**:
We will calculate the cross product \( (\vec{B} - \vec{A}) \times (\vec{C} - \vec{A}) \):
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 3 & -4 \\ -5 & -3 & 4 \end{vmatrix} \]
- The determinant expands to:
\[ \hat{i}(3 \cdot 4 - (-4)(-3)) - \hat{j}(5 \cdot 4 - (-4)(-5)) + \hat{k}(5 \cdot (-3) - 3 \cdot (-5)) \]
- Calculating it:
\[ = \hat{i}(12 - 12) - \hat{j}(20 - 20) + \hat{k}(-15 + 15) = \hat{i}(0) - \hat{j}(0) + \hat{k}(0) = (0, 0, 0) \]

4. **Conclusion**:
Since the cross product is zero, the points are collinear, meaning that an infinite number of planes can pass through these points.

### Part (b): Points (1, 1, 0), (1, 2, 1), (-2, 2, -1)

1. **Define the Points as Vectors**:
Let the points be represented as vectors:
- \( A = (1, 1, 0) \)
- \( B = (1, 2, 1) \)
- \( C = (-2, 2, -1) \)

2. **Calculate Vectors \( \vec{B} - \vec{A} \) and \( \vec{C} - \vec{A} \)**:
- \( \vec{B} - \vec{A} = (1 - 1, 2 - 1, 1 - 0) = (0, 1, 1) \)
- \( \vec{C} - \vec{A} = (-2 - 1, 2 - 1, -1 - 0) = (-3, 1, -1) \)

3. **Cross Product of \( \vec{B} - \vec{A} \) and \( \vec{C} - \vec{A} \)**:
We will calculate the cross product \( (\vec{B} - \vec{A}) \times (\vec{C} - \vec{A}) \):
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ -3 & 1 & -1 \end{vmatrix} \]
- The determinant expands to:
\[ \hat{i}(1 \cdot (-1) - 1 \cdot 1) - \hat{j}(0 \cdot (-1) - 1 \cdot (-3)) + \hat{k}(0 \cdot 1 - 1 \cdot (-3)) \]
- Calculating it:
\[ = \hat{i}(-1 - 1) - \hat{j}(0 + 3) + \hat{k}(0 + 3) = \hat{i}(-2) - \hat{j}(3) + \hat{k}(3) = (-2, -3, 3) \]

4. **Equation of the Plane**:
The equation of the plane can be expressed as:
\[ \vec{r} - \vec{A} \cdot \vec{n} = 0 \]
Where \( \vec{n} = (-2, -3, 3) \) and \( \vec{A} = (1, 1, 0) \):
\[ -2(x - 1) - 3(y - 1) + 3(z - 0) = 0 \]
Simplifying:
\[ -2x + 2 - 3y + 3 + 3z = 0 \implies -2x - 3y + 3z + 5 = 0 \]
Rearranging gives:
\[ 2x + 3y - 3z = 5 \]

### Summary of Solutions:
- **Part (a)**: Infinite number of planes can pass through the points (1, 1, -1), (6, 4, -5), (-4, -2, 3).
- **Part (b)**: The equation of the plane is \( 2x + 3y - 3z = 5 \).