Consider the plane passing through the points `A(2,2,1),B(3,4,2) and C(7,0,6)`
Which one of the following points lies on the plane

Options

A. `(1,0,0)`
B. `(1,0,1)`
C. `(0,0,1)`
D. None of these

Answer

To find which point lies on the plane defined by the points A(2, 2, 1), B(3, 4, 2), and C(7, 0, 6), we will follow these steps:

### Step 1: Determine the equation of the plane
The equation of a plane passing through three non-collinear points A(x1, y1, z1), B(x2, y2, z2), and C(x3, y3, z3) can be determined 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 points A, B, and C:
- A(2, 2, 1) → (x1, y1, z1)
- B(3, 4, 2) → (x2, y2, z2)
- C(7, 0, 6) → (x3, y3, z3)

### Step 2: Set up the determinant
The determinant becomes:

\[ \begin{vmatrix} x - 2 & 1 & 5 \\ y - 2 & 2 & -2 \\ z - 1 & 1 & 5 \\ \end{vmatrix} = 0 \]

Where:
- \(x_2 - x_1 = 3 - 2 = 1\)
- \(x_3 - x_1 = 7 - 2 = 5\)
- \(y_2 - y_1 = 4 - 2 = 2\)
- \(y_3 - y_1 = 0 - 2 = -2\)
- \(z_2 - z_1 = 2 - 1 = 1\)
- \(z_3 - z_1 = 6 - 1 = 5\)

### Step 3: Expand the determinant
Expanding the determinant gives:

\[ (x - 2) \begin{vmatrix} 2 & -2 \\ 1 & 5 \\ \end{vmatrix} - (y - 2) \begin{vmatrix} 1 & 5 \\ 1 & 5 \\ \end{vmatrix} + (z - 1) \begin{vmatrix} 1 & 2 \\ 1 & -2 \\ \end{vmatrix} = 0 \]

Calculating the 2x2 determinants:
1. \(\begin{vmatrix} 2 & -2 \\ 1 & 5 \end{vmatrix} = (2 \cdot 5) - (-2 \cdot 1) = 10 + 2 = 12\)
2. \(\begin{vmatrix} 1 & 5 \\ 1 & 5 \end{vmatrix} = (1 \cdot 5) - (5 \cdot 1) = 5 - 5 = 0\)
3. \(\begin{vmatrix} 1 & 2 \\ 1 & -2 \end{vmatrix} = (1 \cdot -2) - (2 \cdot 1) = -2 - 2 = -4\)

Substituting back into the equation gives:

\[ 12(x - 2) + 0(y - 2) - 4(z - 1) = 0 \]

### Step 4: Simplify the equation
This simplifies to:

\[ 12x - 24 - 4z + 4 = 0 \]

Rearranging gives:

\[ 12x - 4z - 20 = 0 \quad \text{or} \quad 3x - z = \frac{20}{4} = 5 \]

### Step 5: Check which point lies on the plane
Now we check the given points to see which satisfies the equation \(3x - z = 5\).

1. **Point (1, 0, 0)**:
\[ 3(1) - 0 = 3 \quad \text{(not on the plane)} \]

2. **Point (1, 0, 1)**:
\[ 3(1) - 1 = 2 \quad \text{(not on the plane)} \]

3. **Point (0, 0, 1)**:
\[ 3(0) - 1 = -1 \quad \text{(not on the plane)} \]

4. **Point (3, 0, 4)**:
\[ 3(3) - 4 = 9 - 4 = 5 \quad \text{(on the plane)} \]

### Conclusion
The point that lies on the plane is **(3, 0, 4)**.