Given four points `A(2, 1, 0), B(1, 0, 1), C(3, 0, 1) and D(0, 0, 2)`. Point D lies on a line L orthogonal to the plane determined by the points A, B and C.
The equation of the plane ABC is

Options

A. `x+y+z-3=0`
B. `y+z-1=0`
C. `x+z-1=0`
D. `2y+z-1=0`

Explanation

To find the equation of the plane determined by the points A(2, 1, 0), B(1, 0, 1), and C(3, 0, 1), we can follow these steps:

### Step 1: Identify the points
We have the points:
- A(2, 1, 0)
- B(1, 0, 1)
- C(3, 0, 1)

### Step 2: Find vectors AB and AC
We need to find two vectors that lie in the plane. We can find these vectors by subtracting the coordinates of the points:
- Vector **AB** = B - A = (1 - 2, 0 - 1, 1 - 0) = (-1, -1, 1)
- Vector **AC** = C - A = (3 - 2, 0 - 1, 1 - 0) = (1, -1, 1)

### Step 3: Find the normal vector to the plane
The normal vector **N** to the plane can be found by taking the cross product of vectors AB and AC.

Let:
- **AB** = (-1, -1, 1)
- **AC** = (1, -1, 1)

The cross product **N** = **AB** × **AC** is calculated as follows:

\[ \text{N} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & 1 \\ 1 & -1 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ \text{N} = \hat{i}((-1)(1) - (1)(-1)) - \hat{j}((-1)(1) - (1)(1)) + \hat{k}((-1)(-1) - (-1)(1)) \]

\[ = \hat{i}(-1 + 1) - \hat{j}(-1 - 1) + \hat{k}(1 - (-1)) \]

\[ = \hat{i}(0) - \hat{j}(-2) + \hat{k}(2) \]

\[ = 0\hat{i} + 2\hat{j} + 2\hat{k} \]

Thus, the normal vector **N** = (0, 2, 2).

### Step 4: Write the equation of the plane
The general equation of a plane can be written as:
\[ N_x(x - x_0) + N_y(y - y_0) + N_z(z - z_0) = 0 \]
Where \(N_x, N_y, N_z\) are the components of the normal vector and \((x_0, y_0, z_0)\) is a point on the plane (we can use point A).

Substituting the values:
\[ 0(x - 2) + 2(y - 1) + 2(z - 0) = 0 \]
This simplifies to:
\[ 2(y - 1) + 2z = 0 \]
\[ 2y - 2 + 2z = 0 \]
Dividing through by 2:
\[ y + z - 1 = 0 \]

### Final Equation of the Plane
Thus, the equation of the plane determined by points A, B, and C is:
\[ y + z = 1 \]

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