The position vectors of A,B,C are respectively i+2i+ 3k, -i-j+8k, -4i+4j+6k. Then the position vector of the incentre of $△$ ABC is

$\begin{array}{l}a=\left|\overrightarrow{BC}\right|=\sqrt{(-4+1{)}^2+(4+1{)}^2+(6-8{)}^2}=\sqrt{9+25+4}=\sqrt{38}\\ b=\left|\overrightarrow{CA}\right|=\sqrt{(1+4{)}^2+(2-4{)}^2+(3-6{)}^2}=\sqrt{25+4+9}=\sqrt{38}\\ c=\left|\overrightarrow{AB}\right|=\sqrt{(1+1{)}^2+(2+1{)}^2+(3-8{)}^2}=\sqrt{4+9+25}=\sqrt{38}.\text{ The triangle is equilateral. }\\ \therefore \text{ Incentre }=\text{ centroid }=\frac{(\mathbf{i}+2\mathbf{j}+3\mathbf{k})+(-\mathbf{i}-\mathbf{j}+8\mathbf{k})+(-4\mathbf{i}+4\mathbf{j}+6\mathbf{k})}{3}=\frac{1}{3}(-4\mathbf{i}+5\mathbf{j}+17\mathbf{k}).\end{array}$