Even if you specify two corners - (x1,y1,z1) and (x2,y2,z2) - by only specifying (x3,y3,-) you are defining an infinite family of triangles. So, there must be more information you have neglected to give us.
The triangle is defined by 3 points each with an x, y, and z coordinate. Each of those 3 points are known. How would I find the z coordinate of a point that is located inside of the triangle if I knew the x and y coordinates of the point already? Hope that helps clarify it.
Use the plane equation: A*(x-x0)+B*(y-y0)+C*(z-z0)=0 where P(x0,y0,z0)is a point that belongs to this plane and N = [A,B,C] is a wector perpendicular to your plane.
So, P is one of triangles vertices. To find N calculate the cross product of two wectors made of triangle''s vertices. So, if triangles vertices are (x1,y1,z1) (x2,y2,z2) and (x3,y3,z3), N = V1 x V2 (x - cross product) where V1 = [x2-x2,y2-y1,z2-z1], V2 = [x3-x1,y3-y1,z3-z1]
After that insert A,B,C values and known x and y coordinates into plane equation. Rearange it, so you get
z = (C*z0 - A*(x-x0) - B*(y-y0))/C
Note, that C must be different from zero, otherwise this method fails.
Grudzio's answer is basically correct. Just to make things a little clearer though and provide some understanding for those not familiar with the solution method...
The scalar equation of a plane having normal vector n =(a,b,c) that passes through the point Po=(xo,yo,zo) is given by: a(x-xo) + b(y-yo) + c(z-zo) = 0
Given Po and n we can obtain the linear relation of a plane (by collecting terms):
ax + by + cz = d
where d = axo + byo czo
Hence, rearranging gives
z = (d - (ax + by))/c
The stipulation that the plane not be vertical (and hence the normal not horizontal to (x,y) plane) means that c will not be zero. Obviously, if the plane is vertical, then specifying a single (x,y) coordinate cannot uniquely determine a z coordinate on the plane.