求过点(1,1,1)且过直线(x-4)/3=(y+1)/2=z/5 的平面方程

设所求平面方程为 [(x-4)/3-(y+1)/2] + k[(y+1)/2-z/5] = 0 ,将 x = 1 ,y = 1 ,z = 1 代入得 4k-10 = 0,解得 k = 5/2 ,所以,所求平面方程为 [(x-4)/3-(y+1)/2] + 5/2*[(y+1)/2-z/5] = 0 ,化简得 4x+9y-6z-7 = 0 ....