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To solve this problem, we need to determine which points lie in the plane defined by three given points \( A(2, 0, 3) \), \( B(4, 0, 1) \), and \( C(1, -2, 0) \).

### Step 1: Find the equation of the plane

The general form of a plane equation is:
\[
Ax + By + Cz + D = 0
\]
We can find the coefficients \( A \), \( B \), \( C \), and \( D \) using points \( A \), \( B \), and \( C \).

#### Step 2: Find two vectors in the plane
Define two vectors:
- \( \overrightarrow{AB} = B - A = (4 - 2, 0 - 0, 1 - 3) = (2, 0, -2) \)
- \( \overrightarrow{AC} = C - A = (1 - 2, -2 - 0, 0 - 3) = (-1, -2, -3) \)

#### Step 3: Find the normal vector to the plane
The normal vector \( \vec{n} \) to the plane is the cross product of \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \):
\[
\vec{n} = \overrightarrow{AB} \times \overrightarrow{AC}
\]
The cross product is computed as:
\[
\vec{n} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 0 & -2 \\
-1 & -2 & -3
\end{vmatrix} = (0 \times -3 - (-2) \times -2)\mathbf{i} - (2 \times -3 - (-2) \times -1)\mathbf{j} + (2 \times -2 - 0 \times -1)\mathbf{k}
\]
\[
\vec{n} = (-6 - 4)\mathbf{i} - (-6 + 2)\mathbf{j} + (-4)\mathbf{k}
\]
\[
\vec{n} = (-10, 4, -4)
\]
Thus, the normal vector is \( \vec{n} = (-10, 4, -4) \), which gives the equation of the plane as:
\[
-10(x - 2) + 4(y - 0) - 4(z - 3) = 0
\]
Simplifying:
\[
-10x + 20 + 4y - 4z + 12 = 0
\]
\[
-10x + 4y - 4z + 32 = 0
\]
Thus, the equation of the plane is:
\[
10x - 4y + 4z = 32
\]

### Step 4: Check the given points
Now, we check which points satisfy the equation \( 10x - 4y + 4z = 32 \).

1. For \( (-1, 0, 4) \):
   \[
   10(-1) - 4(0) + 4(4) = -10 + 16 = 6 \neq 32
   \]
   This point does not lie on the plane.

2. For \( (0, 0, 5) \):
   \[
   10(0) - 4(0) + 4(5) = 20 \neq 32
   \]
   This point does not lie on the plane.

3. For \( (6, 1, 1) \):
   \[
   10(6) - 4(1) + 4(1) = 60 - 4 + 4 = 60 \neq 32
   \]
   This point does not lie on the plane.

4. For \( (1, 0, 4) \):
   \[
   10(1) - 4(0) + 4(4) = 10 + 16 = 26 \neq 32
   \]
   This point does not lie on the plane.

5. For \( (1, 0, -4) \):
   \[
   10(1) - 4(0) + 4(-4) = 10 - 16 = -6 \neq 32
   \]
   This point does not lie on the plane.

6. For \( (1, 0, 5) \):
   \[
   10(1) - 4(0) + 4(5) = 10 + 20 = 30 \neq 32
   \]
   This point does not lie on the plane.

#### Conclusion:
None of the given points lie in the plane defined by points \( A \), \( B \), and \( C \). Therefore, the correct answer is **Žádný** (None).

Would you like more details on any step or have further questions? Here are some related questions you could explore:

1. How is the cross product of two vectors calculated in detail?
2. What are some alternative methods for finding the equation of a plane in 3D space?
3. How do we interpret the normal vector in the context of a plane's orientation?
4. Can the equation of a plane be expressed in different forms, and how do they compare?
5. How can we check if two planes in 3D space are parallel or intersecting?

**Tip:** To verify if a point lies on a plane, substitute the coordinates into the plane's equation and check if both sides are equal.

V prostoru (3D) máme dány body A[2;0;3], B[4;0;1], C[1;-2;0]. Tyto body určují rovinu v prostoru. Vyberte z následující nabídky všechny body, které leží v rovině ABC: [-1;0;4], Žádný, [0;0;5], [6;1;1], [1;0;4], [1;0;-4], [1;0;5]

Solve a 3D geometry problem by finding the plane equation through points A[2;0;3], B[4;0;1], and C[1;-2;0]. Check which points from the list [-1;0;4], [0;0;5], [6;1;1], and more, lie on the plane. This step-by-step solution involves vector cross products and plane equation verification.

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