How do we show that the surfaces represented by $Pp + Qq = R$ where $p= \frac{\partial z}{\partial x}$ and $q= \frac{\partial z}{\partial y}$ are orthogonal to the surfaces represented by $Pdx + Qdy + Rdz = 0 $
I know that vector $(p, q, -1)$ is perpendicular to vector $(P, Q, R)$ but not able to proceed beyond this.
We can see that for the most general case in which the unknown surface is given in an implicit form, i.e., $S(x,y,z) = C$ (in your case we have $S = z - f(x,y) = 0$), the vector $\nabla S = ( S_x,S_y,S_z)$ (subindices denotes partial differentiation) is an orthogonal vector to $S$.
Then, the PDE whose solution is the surface $S$ is therefore given by:
$$(P,Q,R) \cdot \nabla S = 0, \tag{1}$$ where we note that $(P,Q,R) \perp \nabla S$ (in your case, $(P,Q,R) \perp (p,q,-1))$. If we drew a sketch of the surface $S$ and $\nabla S$, we can see that the vector $(P,Q,R)$ is tangent to $S$ since $(P,Q,R)$ is perpendicular to the normal vector of $S$ (this may sound trivial). Let's write who is the differential of $S$:
$$dS = S_x \, dx + S_y \, dy + S_z \, dz, \tag{2}$$
but $dS = dC = 0$, so we have $(S_x, S_y, S_z)\cdot(dx,dy,dz) = 0$ and therefore, in brief:
$$ \left\{\begin{array}{ll} (P,Q,R) \cdot \nabla S & = 0 \\ \nabla S \cdot (dx,dy,dz) & = 0 \end{array}\right. $$ for which we can deduce that $(P,Q,R) \parallel (dx,dy,dz)$.
Hope this helps.
Cheers!