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

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