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To solve this problem, we need to calculate the volume of the pyramid with vertices \( A_1 \), \( A_2 \), \( A_3 \), and \( A_4 \). The steps involved are as follows:

### Step 1: Determine the Area of the Base
The base of the pyramid is the triangle formed by points \( A_1 \), \( A_2 \), and \( A_3 \). The area of a triangle defined by points \( \mathbf{a} = (x_1, y_1, z_1) \), \( \mathbf{b} = (x_2, y_2, z_2) \), and \( \mathbf{c} = (x_3, y_3, z_3) \) in 3D space can be calculated using the cross product of two vectors in the plane of the triangle.

1. **Vectors forming the base**:
   - \( \vec{A_1A_2} = A_2 - A_1 = (3 - 1, 0 - (-2), 3 - 0) = (2, 2, 3) \)
   - \( \vec{A_1A_3} = A_3 - A_1 = (1 - 1, 4 - (-2), 1 - 0) = (0, 6, 1) \)

2. **Cross product of \( \vec{A_1A_2} \) and \( \vec{A_1A_3} \)**:
   \[
   \vec{A_1A_2} \times \vec{A_1A_3} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 2 & 3 \\ 0 & 6 & 1 \end{vmatrix} = (2 \cdot 1 - 3 \cdot 6) \mathbf{i} - (2 \cdot 1 - 3 \cdot 0) \mathbf{j} + (2 \cdot 6 - 2 \cdot 0) \mathbf{k}
   \]
   Simplifying gives:
   \[
   = (-16, -2, 12)
   \]

3. **Magnitude of the cross product**:
   \[
   \| \vec{A_1A_2} \times \vec{A_1A_3} \| = \sqrt{(-16)^2 + (-2)^2 + 12^2} = \sqrt{256 + 4 + 144} = \sqrt{404}
   \]

4. **Area of the triangle**:
   \[
   \text{Area of } \triangle A_1A_2A_3 = \frac{1}{2} \| \vec{A_1A_2} \times \vec{A_1A_3} \| = \frac{1}{2} \sqrt{404} \approx 10.06
   \]

### Step 2: Calculate the Height from \( A_4 \) to the Base Plane
The height \( h \) of the pyramid is the perpendicular distance from point \( A_4 \) to the plane containing points \( A_1 \), \( A_2 \), and \( A_3 \). The equation of a plane passing through points \( A_1 \), \( A_2 \), and \( A_3 \) can be derived using the normal vector \( \vec{n} = \vec{A_1A_2} \times \vec{A_1A_3} \), which we calculated as \( (-16, -2, 12) \).

The plane equation is:
\[
-16(x - 1) - 2(y + 2) + 12(z - 0) = 0
\]
Simplifying gives:
\[
-16x - 2y + 12z = -18
\]

The general form of the plane equation is:
\[
Ax + By + Cz + D = 0
\]
where \( A = -16 \), \( B = -2 \), \( C = 12 \), and \( D = -18 \).

The distance \( d \) from a point \( (x_0, y_0, z_0) \) to a plane \( Ax + By + Cz + D = 0 \) is:
\[
d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]
Substitute \( A_4 = (2, 1, 7) \):
\[
d = \frac{|-16 \cdot 2 - 2 \cdot 1 + 12 \cdot 7 - 18|}{\sqrt{(-16)^2 + (-2)^2 + 12^2}} = \frac{| -32 - 2 + 84 - 18 |}{\sqrt{404}} = \frac{|32|}{\sqrt{404}} \approx 1.59
\]

### Step 3: Calculate the Volume of the Pyramid
The volume \( V \) of a pyramid with base area \( B \) and height \( h \) is given by:
\[
V = \frac{1}{3} B h
\]
Substitute \( B \approx 10.06 \) and \( h \approx 1.59 \):
\[
V = \frac{1}{3} \times 10.06 \times 1.59 \approx 5.33
\]

### Answer
The volume of the pyramid is approximately \( V \approx 5.33 \) cubic units.

Would you like a detailed explanation of each step or have any specific questions?

---

1. How do we derive the equation of a plane using three points?
2. Why do we use the cross product to find the area of a triangle in 3D space?
3. How is the height calculated in relation to a plane in three-dimensional geometry?
4. Can you explain how the volume formula for a pyramid is derived?
5. What are some practical applications of finding volumes of pyramidal shapes in real life?

**Tip:** For complex geometry problems, breaking down each part (such as base area and height) and solving step-by-step makes it easier to avoid mistakes and understand the solution path.

Вычислить объем пирамиды с вершинами в точках A1, A2, A3, A4, ее высоту, опущенную из вершины A4 на грань (A1, A2, A3), и площадь грани (A1, A2, A3).

Learn how to calculate the volume of a pyramid with vertices A1(1, -2, 0), A2(3, 0, 3), A3(1, 4, 1), and A4(2, 1, 7). This problem includes finding the area of the base, height from A4 to the base plane, and volume of the pyramid. Suitable for students in Grades 10-12.

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