Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0,1) and R be any point (2, 1,6). Then the image of R in the plane P is:

The equation of the plane passing through three points (2, 1, 0), (4, 1, 1) and (5, 0,1)  isx-2y-1z2013-11=0Expand the determinant as below. (x-2)(0+1)-(y-1)(-1)+z(-2)=0⇒x-2+y-1-2z=0⇒x+y-2z=3Suppose that the image of the point (2,1,6)  in the plane x+y-2z=3 is (h,k,l)If the image of a point Px1,y1,z1 in the plane ax+by+cz+d=0 is h,k,l  then h-x1a=k-y1b=l-z1c=-2ax1+by1+ca2+b2Hence, h-21=k-11=l-6-2=-2(-12)6⇒(h,k,l)=(6,5,-2)