I'll try to do it without getting away too much from your point of view.
You have a plane $T : x + 2y - 3z + 3 = 0$, which I can rewrite as
$$
\begin{bmatrix} 1 \\ 2 \\ -3 \end{bmatrix} \begin{bmatrix} x & y & z \end{bmatrix} = -3.
$$
Now you are projecting in the direction $\begin{bmatrix} 1 & 1 & -1 \end{bmatrix}$, so you want to find $\alpha$ such that for arbitrary $x,y,z$ real, we have that
$$
\begin{bmatrix} 1 \\ 2 \\ -3 \end{bmatrix} \left( \begin{bmatrix} x & y & z \end{bmatrix} + \alpha \begin{bmatrix} 1 & 1 & -1 \end{bmatrix} \right)
=
\begin{bmatrix} 1 \\ 2 \\ -3 \end{bmatrix} \begin{bmatrix} x + \alpha & y + \alpha & z - \alpha \end{bmatrix} = -3.
$$
But the last equation can be re-written as
$$
(x+\alpha) + 2(y + \alpha) -3(z - \alpha) = x + 2y - 3z + 6\alpha = -3.
$$
Therefore,
$$
\alpha = \frac{x+2y-3z+3}6
$$
gives you the unique $\alpha$ for which this is possible. If you actually need to compute the algorithm (matrix) which finds the projection as a function of $x$,$y$ and $z$, just compute the vector $[x+\alpha, y+\alpha,z-\alpha]$ and see the matrix there. This would give
$$
\begin{bmatrix} x + \alpha & y + \alpha & z - \alpha \end{bmatrix} = \begin{bmatrix} \frac{7x + 2y - 3z + 3}6 & \frac{x+8y-3z+3}6 & \frac{-x-2y+3z-3}6 \end{bmatrix}
$$
or, written in column form (as standard)
$$
\begin{bmatrix} x + \alpha \\ y + \alpha \\ z - \alpha \end{bmatrix} = \begin{bmatrix} \frac{7x + 2y - 3z + 3}6 \\ \frac{x+8y-3z+3}6 \\ \frac{-x-2y+3z-3}6 \end{bmatrix} = \begin{bmatrix}\frac 76 & \frac 26 & \frac {-3}6 \\ \frac 16 & \frac 86 & \frac {-3}6 \\
\frac {-1}6 & \frac{-2}6 & \frac 36 \end{bmatrix} \begin{bmatrix} x \\ \\ \\ y \\ \\ \\ z \end{bmatrix} + \begin{bmatrix} \frac 36 \\ \frac 36 \\ \frac{-3}6 \end{bmatrix} = P \left( \begin{bmatrix} x \\ \\ \\ y \\ \\ \\ z \end{bmatrix} \right).
$$
where $P$ would be the "projection map".
Hope that helps,
Let's say that your surface is defined by $\vec{r}(\Theta) = [r_0 + s f(\Theta) ] \hat{r}$, where $\Theta$ represents the spherical angles. I'm going to conjecture that the normal vector in radial coordinates is proportional to $\vec{n} = (-s\nabla f,1)$, where the gradient is a 2-vector in the angle space, and the last component is the radial direction.
We can verify this by making sure the normal vector is orthogonal to the surface. Note that a tangent vector to the surface is $(1,s\nabla f)$. This tangent points along the gradient direction. A normal vector orthogonal to this one points along an isocontour, which by construction is $(\hat{\Theta}^\perp,0)$, where $\hat{\Theta}^\perp$ is 90 degree rotation of the unit angular vectors in the tangent plane to the sphere passing through the point. The dot product with both of these is zero; the first is obvious, while for the second, $\nabla f \cdot \hat{\Theta}^\perp = 0$.
TL;DR: In radial coordinates: $(-s\nabla f,1)$, then just normalize and convert to cartesian if needed.
Elaboration: Based on your comment, it seems that your $f(\vec{x})$ is a scalar function of a 3-vector. Then $\nabla f$ is a 3-vector. The notation I use lumps the spherical coordinates $(\phi,\theta)$ into a single abstract 2-vector-like quantity $\Theta$, so that when you say $f(\Theta)$, that means $f$ is a function of only the spherical coordinate angles (and not a function of radius). Then $\nabla f(\Theta)=\nabla_\Theta f(\Theta)$ is a gradient in this 2-dimensional angle space. Your $\nabla f$ needs to be projected onto the surface of a sphere first, since you only ever sample $f(\vec{x})$ for $\lVert \vec{x} \rVert = 1$.
So, to calculate your normals, given a point $\vec{P}(\vec{x}) = [R + s f(\vec{x})]\vec{x}$ with $\vec{x}$ such that $\lVert \vec{x} \rVert = 1$,
- Let $\vec{g}(\vec{x}) = \nabla f(\vec{x})$. Here, $\vec{g}$ is the true gradient of $f$ in 3-space, and you can calculate that in cartesian coordinates, so you get $\vec{g} = (g_x,g_y,g_z)$.
- Project out the radial component of $\vec{g}$ to get $\vec{h}(\vec{x})$. To do this, $\vec{h} = \vec{g} - \frac{\vec{g}\cdot\vec{x}}{\vec{x}\cdot\vec{x}}\vec{x}$. Note that the denominator should be 1. Here, $\vec{h}$ represents the component of $\vec{g}$ that should be tangential to a sphere centered at the origin and passing through $\vec{P}$.
- An outward normal vector to the surface is $\vec{n} = \vec{x} - s\cdot \vec{h}$. Normalize it to get a unit normal vector.
All these computations can be done in Cartesian coordinates, but notice that I never had to resort to referring to the Cartesian components of any vector; the basic operations are vector arithmetic and dot products.
Best Answer
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.