Let the position vectors of the vertices $$\mathrm{A}, \mathrm{B}$$ and $$\mathrm{C}$$ of a triangle be $$2 \hat{i}+2 \hat{j}+\hat{k}, \hat{i}+2 \hat{j}+2 \hat{k}$$ and $$2 \hat{i}+\hat{j}+2 \hat{k}$$ respectively. Let $$l_1, l_2$$ and $$l_3$$ be the lengths of perpendiculars drawn from the ortho center of the triangle on the sides $$\mathrm{AB}, \mathrm{BC}$$ and $$\mathrm{CA}$$ respectively, then $$l_1^2+l_2^2+l_3^2$$ equals:

The position vectors of the vertices $$\mathrm{A}, \mathrm{B}$$ and $$\mathrm{C}$$ of a triangle are $$2 \hat{i}-3 \hat{j}+3 \hat{k}, 2 \hat{i}+2 \hat{j}+3 \hat{k}$$ and $$-\hat{i}+\hat{j}+3 \hat{k}$$ respectively. Let $$l$$ denotes the length of the angle bisector $$\mathrm{AD}$$ of $$\angle \mathrm{BAC}$$ where $$\mathrm{D}$$ is on the line segment $$\mathrm{BC}$$, then $$2 l^2$$ equals :