Find the equation of the plane that passes through the point (-14-2) and contains the line x(t)=4- 3ty(t)=-3+tz(t)=-2-2t. Put equation = 0 please, thanks Solution put (-1 4-2) in the equation of plane -a+4b-2c+d=0 point (4,-3,-2) lies on the plane so the plane ={(4+1,-3-4,-2+2)}=(a,b,c)=(5,-7,0) putting in the above equation c=33 equationn of plane is =5x-7y+33=0.