Question 389306
First determine the equation of the line perpendicular to the given plane and passing through the point (1,-2,4).  Then get the intersection of the line with the given plane.  That intersection point is the midpoint of the segment connecting  (1,-2,4) to the unknown point (a,b,c).

The normal vector to the given plane is <2,-3,-4>.  The symmetric form of the line perpendicular to the given point is then {{{(x-1)/2 = (y+2)/(-3) = (z-4)/(-4)}}}.  Solving for y and z in terms of x, then we get:

{{{y = (-3x-1)/2}}}, and z = -2x+6.  

Putting these two equations into 2x - 3y - 4z + 66 = 0, we get
{{{2x-3((-3x-1)/2) - 4(-2x+6) + 66 = 2x + 9x/2 + 3/2 + 8x-24 + 66 = 0}}}, or {{{29x/2 = -87/2}}} after simplification.
Hence x = -3, y = 4, and z = 12, after substituting back into {{{y = (-3x-1)/2}}} and z = -2x+6.  Thus the intersection point of the normal line and the plane is (-3,4,12). 
To find the reflected point (a,b,c), we use the midpoint formula:  (Recall the given point is (1,-2,4))

{{{(1+a)/2 = -3}}} ----> a = -7
{{{(-2 + b)/2 = 4}}} ----> b = 10
{{{(4+c)/2 = 12}}} ------> c = 20.

Therefore the reflected point is (-7,10,20).