Geometrical Meaning of linear first order partial differential equations? Consider a linear first order differential equation: 
$Pp + qQ = R$ where $p$ and $q$ respectively are derivative of function $f$ wrt $x$ and $y$ respectively and $f(x,y)= z$
Now we can say that at any point say $A$ on $f$ ie the point $A$ is on the graph of the function $f$ in 3D space, we can find the direction cosines of perpendicular to the function $f$ at $A$ as $(p,q,-1)$.
According to Sneddon book however, it also says that the first equation is basically an analytical way of saying that the direction ratios of line perpendicular to normal at $A$ ie tangential to the surface is $(P,Q,R)$ where They are functions of $x, y , z$ not involving any derivative term. 
How is this possible to conclude from the given pde? Can someone please explain this?
Basically I am trying to prove Theorem 2 (See the picture)
 A: OK, so from what we've got so far (i.e., in the text of the question statement):
$p = \dfrac{\partial f(x, y)}{\partial x}, \tag 1$
$q = \dfrac{\partial f(x, y)}{\partial y}, \tag 2$
$z = f(x, y); \tag 3$
then the equation
$Pp + Qq = R \tag 4$
may be written
$P(x, y, z) \dfrac{\partial f(x, y)}{\partial x}  + Q(x, y, z) \dfrac{\partial f(x, y)}{\partial y} = R(x, y, z); \tag 5$
if we define the function
$\phi(x, y, z) = f(x, y) - z, \tag 6$
then it is easy to see that the graph of $f(x, y)$, that is, the set of points $(x, y, z) \in \Bbb R^3$ such that (3) binds, is precisely the $0$-level surface of $\phi(x, y, z)$, since
$f(x, y) - z = \phi(x, y, z) = 0 \Longleftrightarrow z = f(x, y); \tag 7$
we also have the gradient of $\phi(x, y, z)$:
$\nabla \phi(x, y, z) = \left (\dfrac{\partial \phi(x, y, z)}{\partial x}, \dfrac{\partial \phi(x, y, z)}{\partial y}, \dfrac{\partial \phi(x, y, z)}{\partial z} \right ) = \left (\dfrac{\partial f(x, y)}{\partial x}, \dfrac{\partial f(x, y)}{\partial y}, -1 \right )$
$= (p(x, y), q(x, y), -1). \tag 8$
Now it is well-known that $\nabla \phi(x, y, z)$ is normal the the level surfaces of the function $\phi(x, y, z)$; since the graph of $z = f(x, y)$ is the $0$-level surface of $\phi(x, y, z)$, $\nabla \phi(x, y, z) = (p(x. y), q(x, y), -1)$ is normal to the graph of $z = f(x, y)$ at every point $A$ in this graph.  This means in fact that for any vector $X$ tangent to said graph, 
$\nabla \phi(x, y, z) \cdot X = (p(x. y), q(x, y), -1) \cdot X = 0, \tag 9$
and $X$ is tangent to the graph if (9) binds.  It should be observed at this point that $\nabla \phi = (p, q -1)$ is not in general a unit vector; indeed we have
$\Vert \nabla \phi(x, y, z) \Vert = \Vert (p(x, y), q(x, y), -1) \Vert = \sqrt{(p^2(x, y) + q^2(x, y) + 1} \ge 1, \tag{10}$
with equality holding if and only if
$p^2(x, y) + q^2(x, y) = 0 \Longleftrightarrow p(x, y) = q(x, y) = 0; \tag{11}$
we thus see that $\nabla \phi(x, y, z)$ is a unit vector precisely when 
$p(x, y) = \dfrac{\partial f(x, y)}{\partial x} = 0; q(x, y) = \dfrac{\partial f(x, y)}{\partial y}, \tag{12}$
that is, at those points $(x, y) \in \Bbb R^2$ which are critical points of $f(x, y)$; though this fact is perhaps not directly relevant to the present topic, it seems like a worthwhile result worth knowing in its own right.  
We continue.  We may find the direction cosines of $\nabla \phi = (p, q, -1)$ by normalizing it using (10); denoting them by $\alpha_x, \alpha_y, \alpha_z$ we have
$(\alpha_x, \alpha_y, \alpha_z) = \left ( \dfrac{p}{\sqrt{p^2 + q^2 + 1}}, \dfrac{q}{\sqrt{p^2 + q^2 + 1}}, \dfrac{-1}{\sqrt{p^2 + q^2 + 1}} \right ); \tag{13}$
$\alpha = (\alpha_x, \alpha_y, \alpha_z) \tag{14}$
is then of course a unit vector normal to the graph of $z = f(x, y)$.
We return to the equation (5), writing it as
$P(x, y, z) \dfrac{\partial f(x, y)}{\partial x}  + Q(x, y, z) \dfrac{\partial f(x, y)}{\partial y} - R(x, y, z) = 0, \tag{15}$
or
$(P(x, y, z), Q(x, y, z), R(x ,y, z)) \cdot (\dfrac{\partial f(x, y)}{\partial x}, \dfrac{\partial f(x, y)}{\partial y}, -1)$
$= (P(x, y, z), Q(x, y, z), R(x ,y, z)) \cdot (p(x, y), q(x, y), -1) = 0; \tag{16}$
from what we have seen in (9), (16) shows that $(P(x, y, z), Q(x, y, z), R(x ,y, z))$ is tangent to the surface $\phi(x, y, z) = f(x, y) - z = 0$, which is the graph of the equation $z = f(x, y)$.
In the preceding discussion, we normalized the vector field $\nabla (x, y, z) = (p(x, y), q(x, y), -1)$; we may also normalize the vector field $(P, Q, R)$; we have
$\Vert (P, Q, R) \Vert = \sqrt{P^2 + Q^2 + R^2}, \tag{17}$
whence we may, as long as $\Vert (P, Q, R) \Vert \ne 0$, define the vector field $U(x, y, z)$:
$U = \dfrac{(P, Q, R)}{\Vert (P, Q, R) \Vert} = \left ( \dfrac{P}{\sqrt{P^2 + Q^2 + R^2}}, \dfrac{Q}{\sqrt{P^2 + Q^2 + R^2}}, \dfrac{R}{\sqrt{P^2 + Q^2 + R^2}} \right ); \tag{18}$
it is not at all clear, however, whether or not the equation (4)-(5) is any easier to solve when $(P, Q, R)$ is replaced by $U$, in which case it becomes
$ \dfrac{P}{\sqrt{P^2 + Q^2 + R^2}} \dfrac{\partial f(x, y)}{\partial x}  +  \dfrac{Q}{\sqrt{P^2 + Q^2 + R^2}} \dfrac{\partial f(x, y)}{\partial y} =  \dfrac{R}{\sqrt{P^2 + Q^2 + R^2}}; \tag{19}$
(19) may, however, yield certain geometrical insights which are not so readily apparent from the un-nomralized form (5).  But a discussion of this possibility would take us farther afield than time and space permit at this point.
A: To explain this, we need to understand where the geometric interpretation comes from. 
For this it is crucial to know a bit about geometry on manifolds. Here, a vector $X$ is defined as the mapping $X : C(M) \to \mathbb{R}$ which maps functions on the manifold $M$ to the Reals. When we look at $X$ in a chart we see that it looks like a directional derivative acting on a function $X^i\frac{\partial}{\partial x^i}f = X^i\frac{\partial f}{\partial x^i} = X \cdot \nabla f$. Now with your $X = ({P,Q,R})^T$ and $  \nabla f=(p,q,-1)^T$ the equation $X \cdot \nabla f=0$ is your pde and graphically displays the orthogonality of $X$ and $\nabla f$ 
A: Not sure if I understand the question correctly, but : from $Pp+Qq=R$ we rearrange to
$$ (p,q,-1)\cdot (P,Q,R) = 0$$ 
so the vector $\mathbf P =(P,Q,R)$ is normal to $(p,q,-1)$. You noticed that the integral surface $S$ is the set of solutions to $F=0$, 
$$F(x,y,z) = f(x,y)-z , \quad S = \{ (x,y,z) : F(x,y,z) = 0 \}$$ 
with gradient $\nabla F = (p,q,-1)$. Therefore, if I take one point $\mathbf x_0$ in $S$ and evolve by the integral curve equation
$$ \mathbf x' = \mathbf P(\mathbf x), \quad \mathbf x(0) = \mathbf x_0$$ then as $F(\mathbf x_0) = 0$ and
$$ \frac{d}{dt}(F(\mathbf x)) = \nabla F \cdot \mathbf x' =  (p,q,-1)\cdot (P,Q,R) =  0,$$
the resulting curve solves $F(\mathbf x) = 0$. Since $\mathbf x_0$ was arbitrary, any point in $S$ in contained in an integral curve.
