If a hypersurface in $\mathbb{R}^n$ is implicitly defined by the equation $F(x_1, \dots, x_n) = d$ for some $d \in \mathbb{R}$, then you can consider it as a level curve of the function $F$. As you mention, $\nabla F$ is normal to the surface, so, provided you can determine $F$, you can find the normal vector at a point on the surface.
In your situation, the surface is defined by an equation $F(x, y, z) = d$. Furthermore, it is a plane, so $F(x, y, z) = ax + by + cz$, where $a, b, c \in \mathbb{R}$ are yet to be determined. From the figure, we know that $(1, 0, 0)$, $(0, 2, 0)$ and $(0, 0, 3)$ are on the surface, and therefore must satisfy $F(x, y, z) = d$. Substituting in each point, one by one, we get:
\begin{align*}
F(1, 0, 0) &= d \Rightarrow a(1) + b(0) + c(0) = d \Rightarrow a = d,\\
F(0, 2, 0) &= d \Rightarrow a(0) + b(2) + c(0) = d \Rightarrow 2b = d \Rightarrow b = \dfrac{1}{2}d,\\
F(0, 0, 3) &= d \Rightarrow a(0) + b(0) + c(3) = d \Rightarrow 3c = d \Rightarrow c = \dfrac{1}{3}d.
\end{align*}
Now replacing $a, b,$ and $c$ in the equation for $F$, we obtain:
$$dx + \frac{1}{2}dy + \frac{1}{3}dz = d.$$
If $d = 0$, then every element $(x, y, z) \in \mathbb{R}^3$ would satisfy the equation and hence be on the surface. As this is not the case, $d \neq 0$ so we can divide both sides of the above equation by $d$, leaving us with:
$$x + \frac{1}{2}y + \frac{1}{3}z = 1.$$
If you like, you can multiply both sides by $6$ so that all the coefficients are integers, in which case the equation is $6x + 3y + 2z = 6$. Therefore $F(x, y, z) = 6x + 3y + 2z$, so $\nabla F = (\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}) = (6, 3, 2)$ is a vector normal to the surface. As $\|\nabla F\| = \sqrt{6^2 + 3^2 + 2^2} = \sqrt{49} = 7$, $\frac{1}{\|\nabla F\|}\nabla F = \frac{1}{7}(6, 3, 2)$ is a unit normal to the surface.
Note, as this surface is a plane in $\mathbb{R}^3$, you can obtain the normal in a quicker way.
If we can find two linearly independent vectors in the plane, then their cross product will be normal to the plane. We can use the points given to find two such vectors. As $(1, 0, 0)$ and $(0, 2, 0)$ are in the plane, the vector $(0, 2, 0) - (1, 0, 0) = (-1, 2, 0)$ from $(1, 0, 0)$ to $(0, 2, 0)$ is in the plane. Likewise $(0, 3, 0) - (1, 0, 0) = (-1, 0, 3)$ is also in the plane. Clearly these two vectors are linearly independent. Therefore:
$$(-1, 2, 0) \times (-1, 0, 3) = \left| \begin{array}{ccc}
i & j & k \\
-1 & 2 & 0 \\
-1 & 0 & 3 \end{array} \right| = 6i + 3j + 2k = (6, 3, 2),$$
as before.
As for your concerns in higher dimensions, the approach I outlined is the standard approach, except that you usually know $F(x_1, \dots, x_n)$ and $d$, so all that you need to do is calculate $\nabla F = (\frac{\partial F}{\partial x_1}, \dots, \frac{\partial F}{\partial x_n})$. Most of the time $\nabla F$ will not be a constant vector as its components can be functions - in fact, the only time $\nabla F$ is constant is when your surface is a hyperplane in $\mathbb{R}^n$. Either way, $\nabla F$ is a vector which (can) vary from point to point; this is what we call a vector field. More formally, a vector field on a hypersurface $\Sigma$ is a function $\Sigma \to \mathbb{R}^n$; $\nabla F$ is such a function.
These methods are just applications of two different geometric ideas to help you find a normal vector to a surface. I'm sure that you or I could do some variable pushing and prove that they are compatible, but I don't know how enlightening that would be. I think the most important thing is just to understand the geometry behind each of these ideas.
When you have a parametrized surface $r(u,v) = \left< x(u,v), y(u,v), z(u,v) \right>$ and a point $(u_0,v_0)$, you can consider two cross sections of that surface. The functions $$r(u_0,v) = \left< x(u_0,v), y(u_0,v), z(u_0,v) \right>$$
$$r(u,v_0) = \left< x(u,v_0), y(u,v_0), z(u,v_0) \right>$$
define curves in three dimensions which are contained in the plane $r(u,v)$. Convince yourself that a tangent vector to any curve contained in a surface is also tangent to the surface itself. Therefore the vectors
$$ \frac{\partial}{\partial v} r(u_0,v) \big|_{v=v_0} $$
and
$$ \frac{\partial}{\partial u} r(u,v_0) \big|_{u=u_0} $$
are both tangent to the surface at $r(u_0,v_0)$. Convince yourself that if these two vectors were parallel, then $r$ wouldn't look like a curve at this point, rather than a surface, so they should not be parallel. In linear algebra terms, these vectors span the space of tangent vectors. Their cross product will yield a vector which is normal to both of them, and therefore normal to the plane. This is the definition you stated.
The other definition uses the fact that the gradient of a function at a point is perpendicular to the level surface at that point. To understand this, it is helpful to think of the lower dimensional analogy. The gradient of a function $f(x,y)$ (which defines a surface) will be perpendicular to the level curve at any point. This is geometrically obvious if $f(x,y)$ defines a plane. The level curve will be a horizontal line, and the gradient will point in the direction of greatest slope of the plane. The same logic works, in fact, for $f(x,y)$ that is not a plane because the differentiability of $f$ tells us that it behaves like a plane at any given point.
Best Answer
Given a curve a $c$ and with normal $n_0$ at $P_0$ and $P$ outside, the condition is given by
$$P_0\subseteq \{(P+\vec n_o\cdot t) \cap c\}$$
For the circle it suffices that the line is through the center of the circle.
In general, for a generic curve $c$, the problem is hard to solve.