The incentre of the triangle \(\mathrm{ABC}\), whose vertices are \(\mathrm{A}(0,2,1), \mathrm{B}(-2,0,0)\) and \(\mathrm{C}(-2,0,2)\), is

Let \(\bar{a}=2 \hat{j}+\hat{k}, \hat{b}=-2 \hat{i}, \hat{c}=-2 \hat{i}+2 \hat{k}\)
\(\therefore \quad \overline{\mathrm{AB}}=-2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\),
\(\overline{\mathrm{BC}}=2 \hat{\mathrm{k}}\)
\(\begin{aligned} & \overline{\mathrm{AC}}=-2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \Rightarrow|\overline{\mathrm{AB}}|=3,|\overline{\mathrm{BC}}|=2,|\overline{\mathrm{AC}}|=3\end{aligned}\)
Incentre of \(\triangle \mathrm{ABC}\) is given by
\(\frac{|\overline{\mathrm{AB}}| \overline{\mathrm{c}}+|\overline{\mathrm{BC}}| \overline{\mathrm{a}}+|\overline{\mathrm{AC}}| \overline{\mathrm{b}}}{|\overline{\mathrm{AB}}|+|\overline{\mathrm{BC}}|+|\overline{\mathrm{AC}}|}\)
\(\begin{aligned} & =\frac{3(-2 \hat{i}+2 \hat{k})+2(2 \hat{j}+\hat{k})+3(-2 \hat{i})}{3+2+3} \\ & =\frac{-12 \hat{i}+4 \hat{j}+8 \hat{k}}{8} \\ & =-\frac{3}{2} \hat{i}+\frac{1}{2} \hat{j}+\hat{k}\end{aligned}\)