Equation of plane passing through (1,1,0), (-2,2,-1) and (1,2,1) is

Options

A. 2x+3y + 3z=5
B. 2x-3y + 3z=5
C. 2x + 3y - 3z+5=0
D. 2x+3y - 3z=5

Answer

To find the equation of the plane passing through the points \( A(1, 1, 0) \), \( B(-2, 2, -1) \), and \( C(1, 2, 1) \), we can use the determinant method. The general equation of a plane through three points \( A(x_1, y_1, z_1) \), \( B(x_2, y_2, z_2) \), and \( C(x_3, y_3, z_3) \) can be expressed as follows:

\[ \begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0 \]

### Step 1: Identify the points
Let:
- \( A(1, 1, 0) \) → \( (x_1, y_1, z_1) = (1, 1, 0) \)
- \( B(-2, 2, -1) \) → \( (x_2, y_2, z_2) = (-2, 2, -1) \)
- \( C(1, 2, 1) \) → \( (x_3, y_3, z_3) = (1, 2, 1) \)

### Step 2: Set up the determinant
Now we can set up the determinant:

\[ \begin{vmatrix} x - 1 & y - 1 & z - 0 \\ -2 - 1 & 2 - 1 & -1 - 0 \\ 1 - 1 & 2 - 1 & 1 - 0 \end{vmatrix} = 0 \]

This simplifies to:

\[ \begin{vmatrix} x - 1 & y - 1 & z \\ -3 & 1 & -1 \\ 0 & 1 & 1 \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
Now we will calculate the determinant:

\[ = (x - 1) \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} - (y - 1) \begin{vmatrix} -3 & -1 \\ 0 & 1 \end{vmatrix} + z \begin{vmatrix} -3 & 1 \\ 0 & 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} = (1)(1) - (-1)(1) = 1 + 1 = 2 \)

2. \( \begin{vmatrix} -3 & -1 \\ 0 & 1 \end{vmatrix} = (-3)(1) - (-1)(0) = -3 \)

3. \( \begin{vmatrix} -3 & 1 \\ 0 & 1 \end{vmatrix} = (-3)(1) - (1)(0) = -3 \)

Substituting these back into the determinant equation gives:

\[ (x - 1)(2) - (y - 1)(-3) + z(-3) = 0 \]

### Step 4: Expand and simplify
Expanding this results in:

\[ 2x - 2 + 3y - 3 - 3z = 0 \]

Combining like terms:

\[ 2x + 3y - 3z - 5 = 0 \]

### Step 5: Final equation
Rearranging gives us the equation of the plane:

\[ 2x + 3y - 3z = 5 \]

### Conclusion
Thus, the equation of the plane passing through the points \( (1, 1, 0) \), \( (-2, 2, -1) \), and \( (1, 2, 1) \) is:

\[ \boxed{2x + 3y - 3z = 5} \]