Using Gram-Schmidt orthogonalisation process, orthonormalise the liner independent subset (1, 1, 1), (0, 1, 1), (0, 0, 1)\

The first orthonormal vector $u_1$ is obtained by normalizing $v_1$.

$$u_1 = \frac{v_1}{\|v_1\|}$$

First, find the norm of $v_1$:

$$\|v_1\| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$$

Now, calculate $u_1$:

$$u_1 = \frac{1}{\sqrt{3}}(1, 1, 1) = \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$$

To find $u_2$, we first find an orthogonal vector $w_2$ by subtracting the projection of $v_2$ onto $u_1$ from $v_2$.

$$w_2 = v_2 - (v_2 \cdot u_1) u_1$$

First, calculate the dot product $v_2 \cdot u_1$:

$$v_2 \cdot u_1 = (0, 1, 1) \cdot \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) = 0 \cdot \frac{1}{\sqrt{3}} + 1 \cdot \frac{1}{\sqrt{3}} + 1 \cdot \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$$

Now, calculate $w_2$:

$$w_2 = (0, 1, 1) - \frac{2}{\sqrt{3}} \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$$ $$w_2 = (0, 1, 1) - \left(\frac{2}{3}, \frac{2}{3}, \frac{2}{3}\right)$$ $$w_2 = \left(0 - \frac{2}{3}, 1 - \frac{2}{3}, 1 - \frac{2}{3}\right) = \left(-\frac{2}{3}, \frac{1}{3}, \frac{1}{3}\right)$$

Next, normalize $w_2$ to get $u_2$:

$$\|w_2\| = \sqrt{\left(-\frac{2}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2} = \sqrt{\frac{4}{9} + \frac{1}{9} + \frac{1}{9}} = \sqrt{\frac{6}{9}} = \sqrt{\frac{2}{3}}$$ $$u_2 = \frac{w_2}{\|w_2\|} = \frac{1}{\sqrt{\frac{2}{3}}} \left(-\frac{2}{3}, \frac{1}{3}, \frac{1}{3}\right) = \frac{\sqrt{3}}{\sqrt{2}} \left(-\frac{2}{3}, \frac{1}{3}, \frac{1}{3}\right)$$ $$u_2 = \left(-\frac{2\sqrt{3}}{3\sqrt{2}}, \frac{\sqrt{3}}{3\sqrt{2}}, \frac{\sqrt{3}}{3\sqrt{2}}\right) = \left(-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)$$

To find $u_3$, we first find an orthogonal vector $w_3$ by subtracting the projections of $v_3$ onto $u_1$ and $u_2$ from $v_3$.

$$w_3 = v_3 - (v_3 \cdot u_1) u_1 - (v_3 \cdot u_2) u_2$$

First, calculate the dot product $v_3 \cdot u_1$:

$$v_3 \cdot u_1 = (0, 0, 1) \cdot \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) = 0 \cdot \frac{1}{\sqrt{3}} + 0 \cdot \frac{1}{\sqrt{3}} + 1 \cdot \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}}$$

Next, calculate the dot product $v_3 \cdot u_2$:

$$v_3 \cdot u_2 = (0, 0, 1) \cdot \left(-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right) = 0 \cdot \left(-\frac{2}{\sqrt{6}}\right) + 0 \cdot \frac{1}{\sqrt{6}} + 1 \cdot \frac{1}{\sqrt{6}} = \frac{1}{\sqrt{6}}$$

Now, calculate $w_3$:

$$w_3 = (0, 0, 1) - \frac{1}{\sqrt{3}} \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) - \frac{1}{\sqrt{6}} \left(-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)$$ $$w_3 = (0, 0, 1) - \left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right) - \left(-\frac{2}{6}, \frac{1}{6}, \frac{1}{6}\right)$$ $$w_3 = (0, 0, 1) - \left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right) - \left(-\frac{1}{3}, \frac{1}{6}, \frac{1}{6}\right)$$ $$w_3 = \left(0 - \frac{1}{3} - (-\frac{1}{3}), 0 - \frac{1}{3} - \frac{1}{6}, 1 - \frac{1}{3} - \frac{1}{6}\right)$$ $$w_3 = \left(0, -\frac{2}{6} - \frac{1}{6}, \frac{6}{6} - \frac{2}{6} - \frac{1}{6}\right)$$ $$w_3 = \left(0, -\frac{3}{6}, \frac{3}{6}\right) = \left(0, -\frac{1}{2}, \frac{1}{2}\right)$$

Finally, normalize $w_3$ to get $u_3$:

$$\|w_3\| = \sqrt{0^2 + \left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{0 + \frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{2}{4}} = \sqrt{\frac{1}{2}}$$ $$u_3 = \frac{w_3}{\|w_3\|} = \frac{1}{\sqrt{\frac{1}{2}}} \left(0, -\frac{1}{2}, \frac{1}{2}\right) = \sqrt{2} \left(0, -\frac{1}{2}, \frac{1}{2}\right)$$ $$u_3 = \left(0, -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) = \left(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$$

The orthonormal set of vectors is:

$$\boxed{\left\{\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right), \left(-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right), \left(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\right\}}$$