To solve the problem, we need to verify both statements regarding the distance of point D from the plane formed by points A, B, and C, and the volume of the tetrahedron formed by points A, B, C, and D.
### Step 1: Find the vectors AB, AC, and AD
1. **Calculate vector AB:**
\[
\text{A} = (1, -2, 0), \quad \text{B} = (3, 1, 2)
\]
\[
\text{AB} = \text{B} - \text{A} = (3 - 1, 1 - (-2), 2 - 0) = (2, 3, 2)
\]
2. **Calculate vector AC:**
\[
\text{C} = (-1, 1, -1)
\]
\[
\text{AC} = \text{C} - \text{A} = (-1 - 1, 1 - (-2), -1 - 0) = (-2, 3, -1)
\]
3. **Calculate vector AD:**
\[
\text{D} = (1, 0, -1)
\]
\[
\text{AD} = \text{D} - \text{A} = (1 - 1, 0 - (-2), -1 - 0) = (0, 2, -1)
\]
### Step 2: Find the normal vector to the plane formed by points A, B, and C
To find the normal vector \( \mathbf{n} \) to the plane formed by points A, B, and C, we can use the cross product of vectors AB and AC.
1. **Calculate the cross product \( \mathbf{n} = \mathbf{AB} \times \mathbf{AC} \):**
\[
\mathbf{AB} = (2, 3, 2), \quad \mathbf{AC} = (-2, 3, -1)
\]
\[
\mathbf{n} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 3 & 2 \\
-2 & 3 & -1
\end{vmatrix}
\]
\[
= \mathbf{i}(3 \cdot -1 - 2 \cdot 3) - \mathbf{j}(2 \cdot -1 - 2 \cdot -2) + \mathbf{k}(2 \cdot 3 - 3 \cdot -2)
\]
\[
= \mathbf{i}(-3 - 6) - \mathbf{j}(-2 + 4) + \mathbf{k}(6 + 6)
\]
\[
= \mathbf{i}(-9) - \mathbf{j}(2) + \mathbf{k}(12)
\]
\[
= (-9, -2, 12)
\]
### Step 3: Find the equation of the plane
The equation of the plane can be expressed as:
\[
-9(x - 1) - 2(y + 2) + 12(z - 0) = 0
\]
Expanding this gives:
\[
-9x + 9 - 2y - 4 + 12z = 0 \implies -9x - 2y + 12z + 5 = 0
\]
### Step 4: Calculate the distance from point D to the plane
The distance \( d \) from point \( D(1, 0, -1) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[
d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]
Substituting \( A = -9, B = -2, C = 12, D = 5 \) and \( (x_0, y_0, z_0) = (1, 0, -1) \):
\[
d = \frac{|-9(1) - 2(0) + 12(-1) + 5|}{\sqrt{(-9)^2 + (-2)^2 + (12)^2}}
\]
\[
= \frac{|-9 + 0 - 12 + 5|}{\sqrt{81 + 4 + 144}} = \frac{|-16|}{\sqrt{229}} = \frac{16}{\sqrt{229}}
\]
### Step 5: Verify Statement 1
The distance calculated is \( \frac{16}{\sqrt{229}} \), which does not match the given statement \( \frac{8}{\sqrt{229}} \). Therefore, **Statement 1 is false**.
### Step 6: Calculate the volume of the tetrahedron
The volume \( V \) of tetrahedron formed by points A, B, C, and D is given by:
\[
V = \frac{1}{6} | \mathbf{AD} \cdot (\mathbf{AB} \times \mathbf{AC}) |
\]
We already have \( \mathbf{AD} = (0, 2, -1) \) and \( \mathbf{AB} \times \mathbf{AC} = (-9, -2, 12) \).
1. **Calculate the dot product:**
\[
\mathbf{AD} \cdot (\mathbf{AB} \times \mathbf{AC}) = (0, 2, -1) \cdot (-9, -2, 12) = 0 \cdot -9 + 2 \cdot -2 + -1 \cdot 12 = 0 - 4 - 12 = -16
\]
2. **Calculate the volume:**
\[
V = \frac{1}{6} | -16 | = \frac{16}{6} = \frac{8}{3}
\]
### Step 7: Verify Statement 2
The volume calculated is \( \frac{8}{3} \), which does not match the given statement \( \frac{\sqrt{229}}{2} \). Therefore, **Statement 2 is also false**.
### Conclusion
Both statements are incorrect.
