若 $\triangle A B C$ 中, $\angle C=90^{\circ}, A(1,2,-3 k), B(-2,1,0), C(4,0,-2 k)$, 则 $k$ 的值为
$ecause A(1,2,-3 k), B(-2,1,0), C(4,0,-2 k), 	herefore \overrightarrow{A C}=(3,-2, k), \overrightarrow{B C}=(6,-1,-2 k) ecause 	riangle A B C$中, $\angle C=90^{\circ}: \overrightarrow{A C} \cdot \overrightarrow{B C}=(3,-2, k) \cdot(6,-1,-2 k)=18+2-2 k^{2}=0$ 解得 $k= \pm \sqrt{10}$ 故选 $D$.