Probabilities to find $psi$ in the different eigenstates of $hat{L}_z$: how to handle the $r$-dependence?

The wave function of a particle in a spherically symmetric potential $V(r)$ is given by: $$ psi(vec{r})=(x+y+3z)f(r).$$ Determine the probabilities to find $psi$ in the different eigenstates of $hat{L}_z$.

My attempt:

It’s best to rewrite $psi(vec{r})$ in terms of spherical harmonics: $$psi(r,theta,phi) = left(frac {4pi}{3}right)^{1/2}left( -frac{i-1}{sqrt 2} Y_{1,1}(theta,phi)+frac{1+i}{sqrt 2}Y_{1,-1}(theta,phi)+3Y_{1,0}(theta,phi)right)rf(r).$$ This is clearly an eigenstate of $hat{vec{L}}^2$ (with eigenvalue $2hbar^2$) and therefore of $hat{L}_z$. The possible eigenstates of $hat{L}_z$ in which we can observe $psi$ are $Y_{1,1},Y_{1,-1}$ and $Y_{1,0}$, so we need to find $operatorname{Pr}(m=pm 1) $ and $operatorname{Pr}(m=0)$.

Now, $operatorname{Pr}(m=1)=frac{|langle 1,1|psirangle|^2}{langlepsimidpsirangle}$. To find these probability amplitudes, I would use the resolution of the identity w.r.t. $|theta,phirangle=|vec{Omega}rangle$, i.e. $$ langle 1,1midpsirangle=int dvec{Omega} langle 1,1midvec{Omega}ranglelanglevec{Omega}midpsirangle=int_0^{2pi}dphiint_0^{pi}sintheta dtheta Y^{*}_{1,1}(theta,phi)psi(theta,phi).$$ As you can see, the $r$-dependence of $psi$ does not appear in this integral. Can I just take the first factor of $psi$ and use it in this integral? Is this still the same wave function that I’m working with?