The simplest way is to first find the equation of the plane.
So, suppose you are given three points, $ (a_1,b_1,c_1),\quad (a_2,b_2,c_2),\quad (a_3,b_3,c_3).$
I'm first going to assume everything will work out fine; I'll point out the possible problems later.
First, construct two vectors determined by these three points: $\begin{align*} \mathbf{v}_1 &= (a_1,b_1,c_1) - (a_2,b_2,c_2) = (a_1-a_2, b_1-b_2, c_1-c_2).\\ \mathbf{v}_2 &= (a_1,b_1,c_1) - (a_3,b_3,c_3) = (a_1-a_3, b_1-b_3, c_1-c_3). \end{align*}$
Then, compute their cross product: $\mathbf{n} = \mathbf{v}_1\times\mathbf{v}_2 = (r,s,t).$
The plane you want has equation $rx + sy + tz = k$ for some $k$. To find $k$, plug in one of the points you have, say $(a_1,b_1,c_1)$, so you know that $k = ra_1 + sb_1 + tc_1.$
Finally, given the $x$ and $y$ coordinate of a point, you can find the value of $z$ by solving: $z = \frac{1}{t}\left( ra_1 + sb_1 + tc_1 - rx - sy\right).$
What can go wrong?
For three points to determine a unique plane, you need the three points to not be collinear (not lie on the same line). You will find this when you compute the vector $\mathbf{n}$. If $\mathbf{n}=(0,0,0)$, then $\mathbf{v}_1$ and $\mathbf{v}_2$ are parallel, so that means that the three points are collinear and don't determine a unique plane. So you can just test $\mathbf{n}$ to see if it is nonzero before proceedings.
It's possible for there to not be a unique value of $z$ that goes with the given $x$ and $y$. This will happen if $\mathbf{n}$ has the form $\mathbf{n}=(r,s,0)$. Then either the given $x$ and $y$ satisfy the equation you get, in which case every value of $z$ works; or else the given $x$ and $y$ do not satisfy the equation you get and no value of $z$ works.
Example. Suppose you are given $(1,2,3)$, $(1,0,1)$, and $(-2,1,0)$. Then $\begin{align*} \mathbf{v}_1 &= (1,2,3) - (1,0,1) = (0,2,2).\\ \mathbf{v}_2 &= (1,2,3) - (-2,1,0) = (3,1,3). \end{align*}$ Then $\begin{align*} \mathbf{n} &= \mathbf{v}_1\times\mathbf{v}_2 = \left|\begin{array}{rrr} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 0 & 2 & 2\\ 3 & 1 & 3 \end{array}\right|\\ &= ( 6-2, 6- 0, 0-6) \\ &= (4,6,-6). \end{align*}$ So the plane has equation $4x + 6y - 6z = k$. To find $k$, we plug in $(1,2,3)$: $ 4(1) + 6(2) - 6(3) = -2,$ so the plane has equation $4x + 6y - 6z = -2$ or $2x + 3y - 3z = -1.$
Then, given two values of $x$ and $y$, say, $x=7$ and $y=-2$, you plug them in and solve for $z$: $z = \frac{1}{-3}(-1 -2(7) -3(-2)) = \frac{1 + 14 - 6}{3} = 3,$ so the point is $(7,-2,3)$.