Suppose $\boldsymbol R,\boldsymbol P$ are the common position and momentum operators, and $\boldsymbol L=\boldsymbol R\times \boldsymbol P$ is the orbital angular momentum . $\boldsymbol K$ is a linear combination of $\boldsymbol R,\boldsymbol P$, and Jordan product of any of their cross product, e.g. $\frac12(\boldsymbol R\times (\boldsymbol R\times \boldsymbol P)-(\boldsymbol R\times \boldsymbol P)\times \boldsymbol R)$, whose coefficients are scalar functions of $ R^2,P^2, \boldsymbol R\cdot \boldsymbol P$, independent from coordinates, i.e. terms like $\boldsymbol R\cdot \boldsymbol{\hat x}$ are not involved.

Let $\boldsymbol J$ be the collection of three angular momentum operators, then a vector operator $\boldsymbol V$ satisfies

$$ [J_i,V_j]=\text i\hbar\epsilon_{ijk}V_k $$

Question: Can we generally proof that $\boldsymbol K$ satisfies the defnition of vector operators, or is there a counter example? In other words, is $$ [L_i,K_j]=\text i\hbar\epsilon_{ijk}K_k $$ true for all $\boldsymbol K$s? One example is the Runge-Lenz operator of a Hydrogen atom

$$ \boldsymbol B=\frac{\boldsymbol P\times\boldsymbol L-\boldsymbol L\times \boldsymbol P}{2me^2}-\frac{\boldsymbol R}R $$

which is confirmed to be a vector operator.

Note: The key is to justify the fact that an operator directly constructed by a classic vector is indeed a vector operator, so simply referring to the definition can't be the proof.