37.求直线与平面的交点:-|||-(1) _(1):dfrac (x+1)(-2)=dfrac (y+1)(3)=dfrac (z-3)(4) 与 _(1):3x+2y+z=0;-|||-(2)L2: =dfrac {y+3)(-1)=dfrac (z+2)(5) 与平面 +3y-z+3=0;-|||-(2)直线 =dfrac {y)(-4)=dfrac (x-5)(6),dfrac (x)(1)=dfrac (y+2)(2)=dfrac (z-3)(-8) 平行;-|||-(2)通过点 (2,-3,1) 和直线 = 1+5t,-3+t,2t ;-|||-(3)通过两平行直线 dfrac (x-1)(1)=dfrac (y+1)(-2)=dfrac (z-2)(3),dfrac (x)(1)=dfrac (y-1)(-2)=dfrac (z+2)(3);-|||-(4)通过直线 dfrac (x)(2)=dfrac (y)(-1)=dfrac (z-1)(2), 而且平行于直线 dfrac (x-1)(0)=dfrac (y)(1)=dfrac (z)(-1).