Find the reflection of point P(2,4,-6) in the plane x-2y+z=6

If the point P' is a refection of point P in any plane, then both P and P' line on the line perpendicular to that plane. The equation of such line is easy to find: all we need is a point and the direction vector. The direction vector is given by the coefficients near , and the required point is our point P. These lead us to the following equation of a line in 3D: L=(2,4,-6)+k(1,-2,1), where k is scalar parameter. all we need to do now is to find that parameter. We know that, by definition, points P' and P are equidistant from the plane, therefore, the intersection of that line with the plane will be the middle point. After substituting the equation of the line into our equation of the plane and solving for k, we obtain the result that k=3. But because k=3 only gives us our middle point, we need to double it to get P',therefore k=6. Substituting k=6 into our equation of the line gives the point (8,-8,0), which are the desired coordinates.