We have a triangle, defined by the coordinates (x,y,z) $\in \mathbb{R}^3$of its 3 vertices. How do I find a vector, coming from C, so that when we "move" C somewhere on the vector, the length of the altitude (from C to AB) doesn't change? I can kind of imagine it coming towards us from the surface, not perpendicular, but tilted forwards. My problem is finding the vector, given only the coordinates of A, B and C. Any help will be appreciated!
A vector that satisfies your condition should be parallel to $\vec{AB}$ with starting point $C$. So if we have $A(x_1,y_1,z_1)$, $B(x_2,y_2,z_2)$, $C(x_3,y_3,z_3)$ given then $\vec{AB} = (x_2-x_1,y_2-y_1,z_2-z_1)$.
Now, let $D(x,y,z)$ be the point to which we move $C$. Then we need to have $\vec{CD} // \vec{AB}$ which implies $\vec{CD} = k\cdot \vec{AB}$ where $k \in \mathbb{R}-\{0\}$. So we have $$k\cdot(x_2-x_1,y_2-y_1,z_2-z_1) = (x-x_3,y-y_3,z-z_3)$$ $$\implies x = k(x_2-x_1)+x_3,\ y = k(y_2-y_1)+y_3,\ z = k(z_2-z_1)+z_3$$
EDIT: Now I noticed that you may be right about two such vectors. If we take $H$ as the leg of altitude on $AB$, then we can move $C$ on a circle whose center is $H$ and length of radius is $|CH|$.