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### 4. Determine the angles of triangle \( ABC \) with vertices:
\[ A(-1, -2, 4), \ B(-4, -2, 0), \ C(3, -2, 1) \]

#### Solution Approach:
To find the angles of triangle \( ABC \), we can use the vector dot product. The formula for the angle \( \theta \) between two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is:
\[
\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|}
\]
1. First, find vectors \( \overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{BC} \).
2. Use the dot product to find the angles at vertices \( A \), \( B \), and \( C \).

#### Step-by-Step Process:
1. Compute vectors:
   \[
   \overrightarrow{AB} = B - A = (-4 - (-1), -2 - (-2), 0 - 4) = (-3, 0, -4)
   \]
   \[
   \overrightarrow{AC} = C - A = (3 - (-1), -2 - (-2), 1 - 4) = (4, 0, -3)
   \]
   \[
   \overrightarrow{BC} = C - B = (3 - (-4), -2 - (-2), 1 - 0) = (7, 0, 1)
   \]

2. Find magnitudes:
   \[
   |\overrightarrow{AB}| = \sqrt{(-3)^2 + 0^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5
   \]
   \[
   |\overrightarrow{AC}| = \sqrt{4^2 + 0^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5
   \]
   \[
   |\overrightarrow{BC}| = \sqrt{7^2 + 0^2 + 1^2} = \sqrt{49 + 1} = \sqrt{50}
   \]

3. Use the dot product and solve for angles:
   - Angle at \( A \): Use \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \).
   - Angle at \( B \): Use \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \).
   - Angle at \( C \): Use \( \overrightarrow{AC} \) and \( \overrightarrow{BC} \).

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### 5. Determine the distance from point \( M(3, 2, 1) \) to the plane \( \pi \) defined by the points:
\[ M_0(1, 2, 1), \ M_1(3, -4, -2), \ M_2(-3, 2, 3) \]

#### Solution Approach:
1. First, find the normal vector of the plane by calculating the cross product of vectors formed by the points \( M_0, M_1, M_2 \).
2. Use the formula for the distance from a point to a plane:
\[
d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]
where \( (A, B, C) \) is the normal vector of the plane, and \( D \) can be found using one of the points \( M_0, M_1, \) or \( M_2 \).

#### Step-by-Step Process:
1. Compute vectors:
   \[
   \overrightarrow{M_0M_1} = M_1 - M_0 = (3 - 1, -4 - 2, -2 - 1) = (2, -6, -3)
   \]
   \[
   \overrightarrow{M_0M_2} = M_2 - M_0 = (-3 - 1, 2 - 2, 3 - 1) = (-4, 0, 2)
   \]
2. Find the cross product of \( \overrightarrow{M_0M_1} \) and \( \overrightarrow{M_0M_2} \), which gives the normal vector to the plane.
3. Once you have

1. Determine the angles of triangle ABC with vertices A(-1, -2, 4), B(-4, -2, 0), and C(3, -2, 1). 2. Determine the distance from point M(3, 2, 1) to the plane π defined by the points M0(1, 2, 1), M1(3, -4, -2), and M2(-3, 2, 3).

Solve triangle ABC by finding the angles using vector dot products and determine the distance from point M(3, 2, 1) to a plane defined by points M0(1, 2, 1), M1(3, -4, -2), and M2(-3, 2, 3). This problem covers key concepts of analytical geometry, vectors, and plane equations.

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