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.
Q. The perpendicular distance of D from the plane ABC is

Options

A. `sqrt(2)`
B. `(1)/(2)`
C. `2`
D. `(1)/(sqrt(2))`

Answer

To find the perpendicular distance of point D from the plane determined by points A, B, and C, we will follow these steps:

### Step 1: Find the equation of the plane determined by points A, B, and C.

Given points:
- A(2, 1, 0)
- B(1, 0, 1)
- C(3, 0, 1)

To find the equation of the plane, we can use the determinant method. The general form of the equation of a plane is given by:

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

Substituting the coordinates of points A, B, and C:

\[ \begin{vmatrix} x - 2 & y - 1 & z - 0 \\ 1 - 2 & 0 - 1 & 1 - 0 \\ 3 - 2 & 0 - 1 & 1 - 0 \end{vmatrix} = 0 \]

This simplifies to:

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

### Step 2: Calculate the determinant.

Calculating the determinant:

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

Calculating the 2x2 determinants:

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

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

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

Substituting these back into the determinant gives:

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

This simplifies to:

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

Or:

\[ y + z - 1 = 0 \]

### Step 3: Find the perpendicular distance from point D to the plane.

Now we have the equation of the plane:

\[ y + z - 1 = 0 \]

We need to find the perpendicular distance from point D(0, 0, 2) to this plane. The formula for the distance \(D\) from a point \((x_0, y_0, z_0)\) to the plane \(Ax + By + Cz + D = 0\) is given by:

\[ D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

In our case, the coefficients from the plane equation \(y + z - 1 = 0\) are:
- \(A = 0\)
- \(B = 1\)
- \(C = 1\)
- \(D = -1\)

Substituting point D(0, 0, 2):

\[ D = \frac{|0 \cdot 0 + 1 \cdot 0 + 1 \cdot 2 - 1|}{\sqrt{0^2 + 1^2 + 1^2}} = \frac{|2 - 1|}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Final Answer:

The perpendicular distance of point D from the plane ABC is:

\[ \frac{1}{\sqrt{2}} \]