To find the image P' of point P(x_{1},y_{1},z_{1}) in the plane ax+by+cz+d = 0: (1) Write the perpendicular line from P: (x-x_{1})/a = (y-y_{1})/b = (z-z_{1})/c = t. (2) The foot F = (x_{1}+at, y_{1}+bt, z_{1}+ct). (3) Substitute in the plane equation to find t = -(ax_{1}+by_{1}+cz_{1}+d)/($a^{2}$+$b^{2}$+$c^{2}$). (4) For the image, use t' = 2t (double the foot parameter): P' = (x_{1}+2at', y_{1}+2bt', z_{1}+2ct') where t' is the value found in step 3. Equivalently, P' = 2F - P.