Problem: Find the points on the graph of z = 3x^{2}- 4y^{2}at which the vectorn= < 3, 2, 2 > is normal to the tangent plane

I tried finding L(x,y) at point (a,b) and solving for a and b before plugging these values back into the equation for the surface and I got 3a^{2}- 4b^{2}= 6ax - 8by. I knew that the coefficients of the x, y, and z terms of the tangent plane equation would be the x, y, and z values of the normal vector, so I set 6a = 3 and -8b = 2 to get a = 1/2 and b = -1/4 however the answer is (-1/4, 1/8, 1/8).

I know I must be doing something wrong D:

Halp