Since, image of a point (x1,y1,z1) in the plane ax+by+cz+d=0 is ax−x1=by−y1=cz−z1=a2+b2+c2−2(ax1+by1+cz1+d) So, image of point (1,3,4) in the plane 2x−y+z+3=0 is 2x−1=−1y−3=1z−4=4+1+1−2(2−3+4+3) ⇒ 2x−1=−1y−3=1z−4=−2 ⇒x=+1−4,y=3+2,z=−2+4 ⇒x=−3,y=5,z=2 i.e. required point is (−3,5,2)