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