Why gradient vector is perpendicular to the plane I know what gradient vector or $\nabla F$ is and I know how to prove that it is orthogonal to the surface (using calculation - not intuitive).
In a particular case, in which we have a three variable function, I want to know why the gradient vector is perpendicular as mentioned. I mean, not in theoretical terms, but intuitive.
Let $F:\mathbb{R}^3 \to \mathbb{R}$ be a function of three variables.  Say we are looking at the surface defined by $F(x,y,z) = 0$.  By definition, the total derivative of $F$ at $p \in \mathbb{R}^3$ is the best linear approximation to $F$ near $p = (a,b,c)$, in other words 

$$dF\big|_p (\Delta x,\Delta y,\Delta z)= \dfrac{\partial F}{\partial x} \Delta x + \dfrac{\partial F}{\partial y} \Delta y +\dfrac{\partial F}{\partial z} \Delta z$$

and 

$$F(a+\Delta x,b +\Delta y,c + \Delta z) \approx F(p)+dF\big|_p (\Delta x,\Delta y,\Delta z)$$

This is really what the multivariable derivative is all about.  Now we can notice that 

$$dF\big|_p (\Delta x,\Delta y,\Delta z)= \langle\dfrac{\partial F}{\partial x}, \dfrac{\partial F}{\partial y},\dfrac{\partial F}{\partial z}\rangle \cdot \langle \Delta x,\Delta y,\Delta z\rangle$$

Now defining $\nabla F = \langle\dfrac{\partial F}{\partial x}, \dfrac{\partial F}{\partial y},\dfrac{\partial F}{\partial z}\rangle$, we see that 

$$dF\big|_p (\textbf{v})= \nabla F\cdot \textbf{v}$$

This one formula packages a lot of mathematics.

Hopefully this gives you some more intuition about what the gradient does.  In words, to see how much $F(\textbf{p})$ changes when you move away a little bit to $\textbf{p}+\textbf{v}$, just dot product $\nabla F\big|_p$ with $\textbf{v}$.

Armed with this intuitive understanding of the gradient, we can see why it must be perpendicular to the level curves of $F$ quite intuitively.

If $p$ is a point of the surface $F(x,y,z) = 0$, then the tangent vectors $\textbf{v}$ to the surface must satisfy $dF\big|_p(v) = 0$, because moving in the direction of the surface should not change the value of $F$ much since the value of $F$ is constant on the surface.  Translating this into a gradient statement we see that $\nabla F\big|_p \cdot \textbf{v} = 0$ for each tangent vector to the surface.

This just says that $\nabla F\big|_p$ is perpendicular to the tangent plane to the surface at $p$!
Simply put: The gradient at a point p(a,b) is the greatest rate of change of z(a,b). The level curve has constant value z. Therefor for these two "lines" to satisfy there definition, they must be perpendicular to one another.