设A（2,3,1）B(4,1,2)C(6,3,7)D(-5,-4,8）求D到平面ABC的距离谢谢了,

设平面ABC的法向量n=（x,y,z）,∵n AB =0,n AC =0,∴ (x,y,z)(2,-2,1)=0 (x,y,z)(4,0,6)=0 即 2x-2y+z=0 4x+6z=0 x=- 3 2 z y=-z.令z=-2,则n=（3,2,-2）． ∴cos＜n,AD ＞= 3×(-7)+2×(-7)-2×7 32+22+(-2)2 (-7)2+(-7)2+72 ． ∴点D到平面ABC的距离为d,d=| AD ||cos＜n,AD ＞|= 49 17 = 49 17 17 ．