Given vectors \( \vec{a} = (1, 0, 0) \), \( \vec{b} = (1, 1, 0) \), and \( \vec{c} = (1, 1, 1) \), find an orthonormal basis by using the Gram-Schmidt Method.

Step 1 ：Firstly, let's normalize the vector \( \vec{a} \). The magnitude of \( \vec{a} \) is \( ||\vec{a}|| = \sqrt{1^2+0^2+0^2} = 1 \). So, the normalized vector \( \vec{u_1} \) is \( \vec{u_1} = \frac{1}{||\vec{a}||} \cdot \vec{a} = (1, 0, 0) \).

Step 2 ：Next, let's subtract the projection of \( \vec{b} \) onto \( \vec{a} \) from \( \vec{b} \) to get a vector orthogonal to \( \vec{a} \). The projection of \( \vec{b} \) onto \( \vec{a} \) is \( \frac{\vec{b} \cdot \vec{a}}{||\vec{a}||^2} \cdot \vec{a} = (1, 0, 0) \). So, the orthogonal vector \( \vec{v} \) is \( \vec{v} = \vec{b} - \frac{\vec{b} \cdot \vec{a}}{||\vec{a}||^2} \cdot \vec{a} = (0, 1, 0) \). Then, normalize \( \vec{v} \) to get \( \vec{u_2} = \frac{1}{||\vec{v}||} \cdot \vec{v} = (0, 1, 0) \).

Step 3 ：Finally, subtract the projection of \( \vec{c} \) onto \( \vec{a} \) and \( \vec{b} \) from \( \vec{c} \) to get a vector orthogonal to \( \vec{a} \) and \( \vec{b} \). The projection of \( \vec{c} \) onto \( \vec{a} \) and \( \vec{b} \) is \( \frac{\vec{c} \cdot \vec{a}}{||\vec{a}||^2} \cdot \vec{a} + \frac{\vec{c} \cdot \vec{b}}{||\vec{b}||^2} \cdot \vec{b} = (1, 1, 0) \). So, the orthogonal vector \( \vec{w} \) is \( \vec{w} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{||\vec{a}||^2} \cdot \vec{a} - \frac{\vec{c} \cdot \vec{b}}{||\vec{b}||^2} \cdot \vec{b} = (0, 0, 1) \). Then, normalize \( \vec{w} \) to get \( \vec{u_3} = \frac{1}{||\vec{w}||} \cdot \vec{w} = (0, 0, 1) \).