So, DR's of line are (4,5,6) Since, line is parallel to the plane, therefore the normal to plane is perpendicular to the line. ∴ a1a2+b1b2+c1c2=0 Consider the plane is x−2y+z=0 Here, a2=1,b2=−2,c1 So, a1a2b1b2+c1c2 =(4×1)+(5×−2)+(6×1)=4−10+6 =10−10=0 Hence, required plane is x−2y+z=0 So, option (c) is correct.