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.

Options

A. `sqrt(2)`
B. `1//2`
C. 2
D. `1//sqrt2`

Explanation

To solve the problem, we need to find the equation of the plane determined by the points A, B, and C, and then check the position of point D relative to this plane.

### Step-by-Step Solution:

**Step 1: Identify the Points**
- Given points are:
- A(2, 1, 0)
- B(1, 0, 1)
- C(3, 0, 1)
- D(0, 0, 2)

**Step 2: Find Vectors in the Plane**
- We can find two vectors in the plane formed by points A, B, and C:
- 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 \( \vec{n} \) to the plane can be found using the cross product of vectors AB and AC:
\[ \vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & 1 \\ 1 & -1 & 1 \end{vmatrix} \]
- Calculating the determinant:
\[ \vec{n} = \hat{i}((-1)(1) - (1)(-1)) - \hat{j}((-1)(1) - (1)(1)) + \hat{k}((-1)(-1) - (-1)(1)) \]
\[ = \hat{i}(-1 + 1) - \hat{j}(-1 - 1) + \hat{k}(1 + 1) \]
\[ = \hat{i}(0) + \hat{j}(2) + \hat{k}(2) = (0, 2, 2) \]

**Step 4: Equation of the Plane**
- The equation of the plane can be written using the normal vector \( \vec{n} = (0, 2, 2) \) and point A(2, 1, 0):
\[ 0(x - 2) + 2(y - 1) + 2(z - 0) = 0 \]
Simplifying this gives:
\[ 2y + 2z - 2 = 0 \quad \Rightarrow \quad y + z = 1 \]

**Step 5: Check Position of Point D**
- Now we check if point D(0, 0, 2) lies on the plane:
\[ y + z = 0 + 2 = 2 \quad \text{(not equal to 1)} \]
- Therefore, point D does not lie on the plane.

**Step 6: Find the Perpendicular Distance from D to the 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:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
- Here, the plane equation can be rearranged to \( 0x + 2y + 2z - 2 = 0 \), so \( A = 0, B = 2, C = 2, D = -2 \).
- For point D(0, 0, 2):
\[ d = \frac{|0(0) + 2(0) + 2(2) - 2|}{\sqrt{0^2 + 2^2 + 2^2}} = \frac{|0 + 0 + 4 - 2|}{\sqrt{0 + 4 + 4}} = \frac{|2|}{\sqrt{8}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Final Answer:
The perpendicular distance from point D to the plane determined by points A, B, and C is \( \frac{1}{\sqrt{2}} \).