To solve the problem, we need to find the position vector of the vertex \( A_1 \) of the right triangular prism \( ABCA_1B_1C_1 \) given the volume and the position vectors of the vertices of the base triangle \( ABC \).
### Step 1: Calculate the area of triangle \( ABC \)
The vertices of triangle \( ABC \) are given as:
- \( A(1, 0, 1) \)
- \( B(2, 0, 0) \)
- \( C(0, 1, 0) \)
To find the area of triangle \( ABC \), we can use the formula for the area of a triangle given by its vertices:
\[
\text{Area} = \frac{1}{2} \left| \vec{AB} \times \vec{AC} \right|
\]
First, we need to find the vectors \( \vec{AB} \) and \( \vec{AC} \):
\[
\vec{AB} = B - A = (2 - 1, 0 - 0, 0 - 1) = (1, 0, -1)
\]
\[
\vec{AC} = C - A = (0 - 1, 1 - 0, 0 - 1) = (-1, 1, -1)
\]
Now, we can calculate the cross product \( \vec{AB} \times \vec{AC} \):
\[
\vec{AB} \times \vec{AC} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 0 & -1 \\
-1 & 1 & -1
\end{vmatrix}
\]
Calculating the determinant:
\[
= \hat{i} \left(0 \cdot (-1) - (-1) \cdot 1\right) - \hat{j} \left(1 \cdot (-1) - (-1) \cdot -1\right) + \hat{k} \left(1 \cdot 1 - 0 \cdot -1\right)
\]
\[
= \hat{i} (0 + 1) - \hat{j} (-1 - 1) + \hat{k} (1 - 0)
\]
\[
= \hat{i} (1) + \hat{j} (2) + \hat{k} (1) = (1, 2, 1)
\]
Now, we find the magnitude of the cross product:
\[
\left| \vec{AB} \times \vec{AC} \right| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}
\]
Thus, the area of triangle \( ABC \) is:
\[
\text{Area} = \frac{1}{2} \left| \vec{AB} \times \vec{AC} \right| = \frac{1}{2} \sqrt{6}
\]
### Step 2: Relate volume to area and height
The volume \( V \) of the prism is given by:
\[
V = \text{Area of base} \times \text{Height}
\]
We know the volume is 3:
\[
3 = \frac{1}{2} \sqrt{6} \cdot h
\]
Solving for \( h \):
\[
h = \frac{3 \cdot 2}{\sqrt{6}} = \frac{6}{\sqrt{6}} = \sqrt{6}
\]
### Step 3: Find the position vector of vertex \( A_1 \)
Let the position vector of \( A_1 \) be \( A_1(x, y, z) \). Since \( A_1 \) is directly above point \( A \) at height \( h \), we have:
\[
A_1 = A + h \cdot \hat{k} = (1, 0, 1 + \sqrt{6}) = (1, 0, 1 + \sqrt{6})
\]
### Step 4: Check the conditions for the right prism
The position vector \( A_1 \) must also satisfy the condition that the vector \( A_1 - A \) is perpendicular to the base \( ABC \). We can check this by ensuring that:
\[
(A_1 - A) \cdot \vec{AB} = 0 \quad \text{and} \quad (A_1 - A) \cdot \vec{AC} = 0
\]
Calculating \( A_1 - A \):
\[
A_1 - A = (1, 0, 1 + \sqrt{6}) - (1, 0, 1) = (0, 0, \sqrt{6})
\]
Now, checking the dot products:
\[
(0, 0, \sqrt{6}) \cdot (1, 0, -1) = 0 \quad \text{(satisfied)}
\]
\[
(0, 0, \sqrt{6}) \cdot (-1, 1, -1) = 0 \quad \text{(satisfied)}
\]
### Conclusion
The position vector of vertex \( A_1 \) can be:
\[
A_1 = (1, 0, 1 + \sqrt{6})
\]
