To solve the problem, we need to find the volume of the tetrahedron formed by the origin O and the points A, B, and C where the plane intersects the axes. Let's go through the solution step by step.
### Step 1: Find the equation of the plane
The plane passes through the point \( (1, 1, 1) \) and is parallel to the vectors \( \vec{b} = (1, 0, -1) \) and \( \vec{c} = (-1, 1, 0) \).
To find the equation of the plane, we can use the fact that if a plane passes through a point \( (x_0, y_0, z_0) \) and is parallel to vectors \( \vec{b} \) and \( \vec{c} \), then the normal vector \( \vec{n} \) to the plane can be found using the cross product of \( \vec{b} \) and \( \vec{c} \).
Calculating the cross product:
\[
\vec{n} = \vec{b} \times \vec{c} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 0 & -1 \\
-1 & 1 & 0
\end{vmatrix}
\]
Calculating the determinant:
\[
\vec{n} = \hat{i}(0 \cdot 0 - (-1) \cdot 1) - \hat{j}(1 \cdot 0 - (-1) \cdot -1) + \hat{k}(1 \cdot 1 - 0 \cdot -1)
\]
\[
= \hat{i}(1) - \hat{j}(1) + \hat{k}(1) = (1, -1, 1)
\]
Now, using the point-normal form of the plane equation:
\[
1(x - 1) - 1(y - 1) + 1(z - 1) = 0
\]
This simplifies to:
\[
x - y + z = 1
\]
### Step 2: Find the intercepts with the axes
To find the points A, B, and C where the plane intersects the axes:
1. **X-axis (y = 0, z = 0)**:
\[
x - 0 + 0 = 1 \implies x = 1 \implies A(1, 0, 0)
\]
2. **Y-axis (x = 0, z = 0)**:
\[
0 - y + 0 = 1 \implies -y = 1 \implies y = -1 \implies B(0, -1, 0)
\]
3. **Z-axis (x = 0, y = 0)**:
\[
0 - 0 + z = 1 \implies z = 1 \implies C(0, 0, 1)
\]
### Step 3: Calculate the volume of tetrahedron OABC
The volume \( V \) of tetrahedron formed by points O(0, 0, 0), A(1, 0, 0), B(0, -1, 0), and C(0, 0, 1) is given by the formula:
\[
V = \frac{1}{6} \left| \begin{vmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1
\end{vmatrix} \right|
\]
Calculating the determinant:
\[
= \frac{1}{6} \left| 1 \cdot (-1) \cdot 1 + 0 + 0 - 0 - 0 - 0 \right| = \frac{1}{6} \left| -1 \right| = \frac{1}{6}
\]
### Final Volume Calculation
Thus, the volume of the tetrahedron OABC is:
\[
V = \frac{1}{6} \text{ cubic units}
\]
