Say you have a triangle with the following verticies on a 2D plane:
Tri = [ (0,0), (0,1), (0.5, 0.5) ]
You wish to move V3 from (0.5,0.5) to (2,2), but you can not directly interact with this point.
Is it possible to apply a scaling transformation to the triangle such that V3 will move upwards, but V1 and V2 remain unchanged? It is okay if you have to apply an additional translation to the triangle to recenter it after the scale, but the positions of V1 and V2 must be the same as the original.
If it's possible, does that hold true to moving all 3 verticies at once? IE we must apply a transformation such that the new Tri values are:
TriNew = [ (-1,0), (3,3), (2,0) ]
The class of transformations you are considering are called dilations. A dilation is by definition the composition of a homothety (of the form $\vec{x}_i' = \lambda(\vec{x}_i-\vec{x}_0)+\vec{x}_0$, where the constant $\vec{x}_0$ "center" is the unique vector fixed under a nontrivial transformation and the constant $\lambda$ "ratio" is responsible for the magnitude of scaling and possible reflection if $\lambda<0$ about the center) with a translation ($\vec{x}_i'=\vec{x}_i+\Delta{\vec{x}}$ for constant $\Delta{\vec{x}}$). Therefore, the general dilation is of the form $\vec{x}'_i = \lambda\vec{x}_i+\vec{c}$ for constant $\lambda,\vec{c}$.
It is interesting to note that dilations are closed under function composition, by which I mean for arbitrary dilations $D_1:\vec{x}_i \mapsto \lambda_1\vec{x}_i+\vec{c}_1$ and $D_2:\vec{x}_i \mapsto \lambda_2\vec{x}_i+\vec{c}_2$, the composition $$D_1 \circ D_2:\vec{x}_i \mapsto \lambda_1(\lambda_2\vec{x}_i+\vec{c}_2)+\vec{c}_1=(\lambda_1\lambda_2)\vec{x}_i+(\lambda_1 \vec{c}_2 + \vec{c}_1)$$ is itself a dilation with $\lambda = \lambda_1 \lambda_2, \vec{c}=\lambda_1 \vec{c}_2 + \vec{c}_1$. In fact, the set of dilations is a group under the operation of function composition (the identity transformation has $\lambda=1, \vec{c}=\vec{0}$ and the inverse transformation of $\vec{x}_i' = \lambda \vec{x}_i + \vec{c}$ is $\vec{x}_i' = \frac{1}{\lambda}\vec{x}_i-\frac{1}{\lambda} \vec{c}$ for $\lambda \neq 0$ and $\vec{x}_i' = \vec{x}_i - \vec{c}$ for $\lambda = 0$). I included this discussion so that we could understand why a sequence of dilations is always equivalent to one well-chosen dilation. Therefore, in determining whether a sequence of dilations is capable of fixing two points while moving another, we only need to consider if a single dilation can do this.
Suppose that the dilation $\vec{x}_i'=\lambda \vec{x}_i + \vec{c}$ fixes $\vec{x}_1, \vec{x}_2$. This means that $\vec{x}_1=\lambda \vec{x}_1 + \vec{c}$ and $\vec{x}_2=\lambda \vec{x}_2 + \vec{c}$. Subtracting these two vector equations yields $\vec{x}_1-\vec{x}_2 = \lambda(\vec{x}_1-\vec{x}_2) \iff (\lambda -1)(\vec{x}_1-\vec{x}_2) = \vec{0}$. Since $\vec{x}_1$ and $\vec{x}_2$ refer to distinct points, we require $\lambda =1$. Returning to either of the two original vector equations, it is clear we also need $\vec{c}=\vec{0}$. We have therefore shown that any dilation that fixes two points must be the identity transformation, which fixes all points. We conclude that it is impossible for a sequence of dilations to fix two points while performing a non-trivial transformation on another point.