Find the equation of the plane passing through the points `(2,1,0),(5,0,1)` and `(4,1,1)`.

Explanation

To find the equation of the plane passing through the points \( A(2, 1, 0) \), \( B(5, 0, 1) \), and \( C(4, 1, 1) \), we can use the determinant method. The equation of the plane can be represented using the determinant of a matrix formed by the coordinates of the points.

### Step-by-Step Solution:

1. **Identify the Points:**
Let the points be:
- \( A(x_1, y_1, z_1) = (2, 1, 0) \)
- \( B(x_2, y_2, z_2) = (5, 0, 1) \)
- \( C(x_3, y_3, z_3) = (4, 1, 1) \)

2. **Set Up the Determinant:**
The equation of the plane can be found using the determinant:
\[ \begin{vmatrix} x - x_1 & x_2 - x_1 & x_3 - x_1 \\ y - y_1 & y_2 - y_1 & y_3 - y_1 \\ z - z_1 & z_2 - z_1 & z_3 - z_1 \end{vmatrix} = 0 \]
Substituting the coordinates of the points:
\[ \begin{vmatrix} x - 2 & 5 - 2 & 4 - 2 \\ y - 1 & 0 - 1 & 1 - 1 \\ z - 0 & 1 - 0 & 1 - 0 \end{vmatrix} = 0 \]
This simplifies to:
\[ \begin{vmatrix} x - 2 & 3 & 2 \\ y - 1 & -1 & 0 \\ z & 1 & 1 \end{vmatrix} = 0 \]

3. **Calculate the Determinant:**
We can expand this determinant. Using the first row for expansion:
\[ = (x - 2) \begin{vmatrix} -1 & 0 \\ 1 & 1 \end{vmatrix} - 3 \begin{vmatrix} y - 1 & 0 \\ z & 1 \end{vmatrix} + 2 \begin{vmatrix} y - 1 & -1 \\ z & 1 \end{vmatrix} \]

Now calculating these 2x2 determinants:
\[ = (x - 2)(-1 \cdot 1 - 0 \cdot 1) - 3((y - 1) \cdot 1 - 0 \cdot z) + 2((y - 1) \cdot 1 - (-1) \cdot z) \]
\[ = -(x - 2) - 3(y - 1) + 2(y - 1 + z) \]
\[ = -x + 2 - 3y + 3 + 2y - 2 + 2z \]
\[ = -x - y + 2z + 3 = 0 \]

4. **Rearranging the Equation:**
Rearranging gives us:
\[ x + y - 2z - 3 = 0 \]
Thus, the equation of the plane is:
\[ x + y - 2z = 3 \]

### Final Answer:
The equation of the plane passing through the points \( (2, 1, 0) \), \( (5, 0, 1) \), and \( (4, 1, 1) \) is:
\[ x + y - 2z = 3 \]