1.先求出过点与直线的垂直的平面：法向量为n=|i j k 1 1 -1 2 -1 1| =(0,-3,-3)=-3(0,1,1) 所以平面为：（y+1）+(z-2)=0 y+z-1=0 2.求出交点 x+y-z+1=0,2x-y+z-4=0 y+z-1=0 解得 x=1 y=-1/2 z=3/2 3.距离 d=√(3-1)²+(-1+1/2)²+(2-3/2)²=(3√2)/2