$$W:=~\text{linear subspace of}~\mathbb{R}^{n}$$
$$\left\{\boldsymbol{a_{1}},\boldsymbol{a_{2}},...\boldsymbol{a_{m}}\right\}:=~\text{basis of this linear subspace}$$
$$\boldsymbol{b}_{1}:=\frac{1}{\Vert\boldsymbol{a}_{1}\Vert}\boldsymbol{a}_{1}$$
$$k\in\mathbb{N}_{\geq2}$$
$$\boldsymbol{b}_{k}':=\boldsymbol{a}_{k}-\underbrace{\left(\boldsymbol{a}_{k},\boldsymbol{b}_{k-1}\right)}_{~\text{inner product}}\boldsymbol{b}_{k-1}-\cdot\cdot\cdot-\left(\boldsymbol{a}_{k},\boldsymbol{b}_{1}\right)\boldsymbol{b}_{1} $$
$$=\boldsymbol{a}_{k}-\sum_{i=1}^{k-1}\left(\boldsymbol{a}_{k},\boldsymbol{b}_{i}\right)\boldsymbol{b}_{i}~~\leftarrow~~~\text{I summarized terms to sigma stuff}$$
$$\boldsymbol{b}_{k}=\frac{1}{\Vert\boldsymbol{\boldsymbol{b}_{k}'\Vert}}\boldsymbol{b}_{k}'$$
$$\left\{\boldsymbol{b}_{1},...,\boldsymbol{b}_{m}\right\}~~\leftarrow~~~\text{orthonormal basis of the linear subspace}$$
$$\boldsymbol{a}_{1}:=\left(2,1,1\right)$$
$$\boldsymbol{a}_{2}:=\left(1,3,2\right)$$
My calculations are as below.
$$\boldsymbol{b}_{1}=\frac{\boldsymbol{a}_{1}}{\Vert\boldsymbol{a}_{1}\Vert}$$
$$=\frac{\left(2,1,1\right)}{\sqrt{2^{2}+1^{2}+1^{2}}}$$
$$=\frac{\left(2,1,1\right)}{\sqrt{4+1+1}}$$
$$=\frac{\left(2,1,1\right)}{\sqrt{6}}$$
$$=\frac{1}{\sqrt{6}}\boldsymbol{a}_{1}$$
$$k=2~~\text{is set}$$
$$\boldsymbol{b}_{2}'=\boldsymbol{a}_{2}-\sum_{i=1}^{2-1}\left(\boldsymbol{a}_{2},\boldsymbol{b}_{i}\right)\boldsymbol{b}_{i}$$
$$=\boldsymbol{a}_{2}-\sum_{i=1}^{1}\left(\boldsymbol{a}_{2},\boldsymbol{b}_{i}\right)\boldsymbol{b}_{i}$$
$$=\boldsymbol{a}_{2}-\left(\boldsymbol{a}_{2},\boldsymbol{b}_{1}\right)\boldsymbol{b}_{1}$$
$$\left(\boldsymbol{a}_{2},\boldsymbol{b}_{1}\right)=\left(1,3,2\right)\cdot\left(\frac{1}{\sqrt{6}}\boldsymbol{a}_{1}\right)$$
$$=\frac{1}{\sqrt{6}}\left(1,3,2\right)\cdot\left(2,1,1\right)$$
$$=\frac{1}{\sqrt{6}}\left\{1\cdot 2+3\cdot 1+2\cdot 1\right\}$$
$$=\frac{1}{\sqrt{6}}\left\{2+3+2\right\}$$
$$=\frac{1}{\sqrt{6}}\left(7\right)$$
$$=\frac{7}{\sqrt{6}}$$
$$\boldsymbol{b}_{2}'=\boldsymbol{a}_{2}-\frac{7}{\sqrt{6}}\left(\frac{\boldsymbol{a}_{1}}{\sqrt{6}}\right)$$
$$=\left(1,3,2\right)-\frac{7}{6}\left(2,1,1\right)$$
$$=\frac{6}{6}\left(1,3,2\right)-\frac{7}{6}\left(2,1,1\right)$$
$$=\frac{1}{6}\left\{6\left(1,3,2\right)-7\left(2,1,1\right)\right\}$$
$$=\frac{1}{6}\left\{\left(6,18,12\right)-\left(14,7,7\right)\right\}$$
$$=\frac{1}{6}\left(-12,11,5\right)$$
$$\Vert\boldsymbol{b}_{2}'\Vert=\sqrt{\left(\frac{1}{6^{2}}\right)\left\{\left(-12\right)^{2}+\left(11\right)^{2}+\left(5\right)^{2}\right\}}$$
$$=\frac{1}{6}\sqrt{144+121+25}$$
$$=\frac{1}{6}\sqrt{265+25}$$
$$=\frac{1}{6}\sqrt{290}$$
$$\therefore~~~\boldsymbol{b}_{2}=\frac{1}{\Vert\boldsymbol{b}_{2}'\Vert}\boldsymbol{b}_{2}'$$
$$=\frac{6}{\sqrt{290}}\frac{1}{6}\left(-12,11,5\right)$$
$$=\frac{1}{\sqrt{290}}\left(-12,11,5\right)$$
The official answer says that the value inside the root is$~210~$
Where I've made mistake(s)?
I think there is a mistake in the line when you write
$$ \frac{1}{6}\left\{ \left(6,18,12\right) - \left(14,7,7\right) \right\} $$
$$ = \frac{1}{6} \left(-12,11,5\right) $$