Answer
$B'=\{(1,0,0),(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}),(0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})\}$ is orthonormal basis for $R^3$.
Work Step by Step
Let $B=\{ v_1=(1,0,0),v_2=(1,1,1), v_3=(1,1,-1) \}$.
Applying the Gram-Schmidt orthonormalization process produces
$$w_1=v_1=(1,0,0)$$
$$w_2=v_2- \frac{v_2\cdot w_1}{w_1\cdot w_1}w_1=(1,1,1)- (1,0,0)=(0,1,1)$$
$$w_3=v_3- \frac{v_3\cdot w_1}{w_1\cdot w_1}w_1- \frac{v_3\cdot w_2}{w_2\cdot w_2}w_2\\=(1,1,-1)-(1,0,0)-0(0,1,1)=(0,1,-1).$$
Normalizing $w_1$, $w_2$ and $w_3$ produces the orthonormal set
$$u_1=\frac{w_1}{\|w_1\|}=(1,0,0) $$
$$u_2=\frac{w_2}{\|w_2\|}=\frac{1}{2}(0,1,1)=(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$$
$$u_3=\frac{w_3}{\|w_3\|}=\frac{1}{2}(0,1,-1)=(0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$$
Hence, $B'=\{(1,0,0),(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}),(0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})\}$ is orthonormal basis for $R^3$.
